Calculate the value of the sample variance. Round, №17886104, 25.12.2022 02:37

Mathematics

Calculate the value of the sample variance. Round your answer to one decimal place. 9_5,9_5,2,9_5

Step-by-step answer

P Answered by PhD

s^2 = 0.01

Step-by-step explanation:

Given

Values: 9/5, 9/5, 2, 9/5

Required

Calculate the sample variance

Sample variance is calculated using:

$s^2 = \frac{\sum (x_i - \overline x)^2}{n - 1}$

First, we calculate the mean

$\overline x = \frac{\sum x}{n}$

$\overline x = \frac{9/5 + 9/5 + 2 + 9/5}{4}$

$\overline x = \frac{7.4}{4}$

$\overline x = 1.85$

$s^2 = \frac{\sum (x_i - \overline x)^2}{n - 1}$ becomes

$s^2 = \frac{(9/5 - 1.85)^2+(9/5 - 1.85)^2+(2 - 1.85)^2+(9/5 - 1.85)^2}{4 - 1}$

$s^2 = \frac{(-0.05)^2+(-0.05)^2+(0.15)^2+(-0.05)^2}{4 - 1}$

$s^2 = \frac{0.0025+0.0025+0.0225+0.0025}{3}$

$s^2 = \frac{0.03}{3}$

s^2 = 0.01

Hence, the variance is 0.01

Mathematics

The null and alternate hypotheses are: H0 : mu1 = mu2 H1 : mu1 not equal mu2 A random sample of 9 observations from one population revealed a sample mean of 22 and a sample deviation of 4.7. A random sample of 9 observations from another population revealed a sample mean of 26 and a sample standard deviation of 3.6. At the .10 significance level, is there a difference between the population means? (a) State the decision rule. (Negative values should be indicated by a minus sign. Round your answer to 3 decimal places.) The decision rule is to reject

Step-by-step answer

P Answered by Specialist

There is a difference between the population means at 0.10 significance level.

(a) The decision rule is to reject H0 if the test statistic (t) falls outside the region bounded by the critical values.

(b) Pooled estimate of the population variance is 17.525.

(c) Test statistic is -2.027

(d) Reject H0

Step-by-step explanation:

At 0.10 significance level, there is a difference between the population means because the null hypothesis is rejected.

(a) The decision rule is to reject H0 if the test statistic (t) falls outside the region bounded by the critical values.

(b) n1 = 9

s1 = 4.7

s1^2 = 4.7^2 = 22.09

n2 = 9

s2 = 3.6

s2^2 = 3.6^2 = 12.96

Pooled variance = [(n1-1)s1^2 + (n2-1)s2^2] ÷ (n1+n2-2) = [(9-1)22.09 + (9-1)12.96] ÷ (9+9-2) = 280.4 ÷ 16 = 17.525

(c) Test statistic (t) = (mean1 - mean2) ÷ sqrt[pooled variance (1/n1 + 1/n2)] = (22 - 26) ÷ sqrt[17.525(1/9 + 1/9)] = -4 ÷ 1.973 = -2.027

(d) degree of freedom = n1+n2-2 = 9+9-2 = 16

significance level = 0.10 = 10%

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.

critical values corresponding to 16 degrees of freedom and 10% significance level are -1.746 and 1.746

My decision rule:

Reject H0 because the test statistic -2.027 falls outside the region bounded by the critical values -1.746 and 1.746.

Mathematics

The null and alternate hypotheses are: H0 : mu1 = mu2 H1 : mu1 not equal mu2 A random sample of 9 observations from one population revealed a sample mean of 22 and a sample deviation of 4.7. A random sample of 9 observations from another population revealed a sample mean of 26 and a sample standard deviation of 3.6. At the .10 significance level, is there a difference between the population means? (a) State the decision rule. (Negative values should be indicated by a minus sign. Round your answer to 3 decimal places.) The decision rule is to reject

Step-by-step answer

P Answered by Specialist

There is a difference between the population means at 0.10 significance level.

(a) The decision rule is to reject H0 if the test statistic (t) falls outside the region bounded by the critical values.

(b) Pooled estimate of the population variance is 17.525.

(c) Test statistic is -2.027

(d) Reject H0

Step-by-step explanation:

At 0.10 significance level, there is a difference between the population means because the null hypothesis is rejected.

(a) The decision rule is to reject H0 if the test statistic (t) falls outside the region bounded by the critical values.

(b) n1 = 9

s1 = 4.7

s1^2 = 4.7^2 = 22.09

n2 = 9

s2 = 3.6

s2^2 = 3.6^2 = 12.96

Pooled variance = [(n1-1)s1^2 + (n2-1)s2^2] ÷ (n1+n2-2) = [(9-1)22.09 + (9-1)12.96] ÷ (9+9-2) = 280.4 ÷ 16 = 17.525

(c) Test statistic (t) = (mean1 - mean2) ÷ sqrt[pooled variance (1/n1 + 1/n2)] = (22 - 26) ÷ sqrt[17.525(1/9 + 1/9)] = -4 ÷ 1.973 = -2.027

(d) degree of freedom = n1+n2-2 = 9+9-2 = 16

significance level = 0.10 = 10%

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.

critical values corresponding to 16 degrees of freedom and 10% significance level are -1.746 and 1.746

My decision rule:

Reject H0 because the test statistic -2.027 falls outside the region bounded by the critical values -1.746 and 1.746.

Mathematics

An article reported the following data on oxidation-induction time (min) for various commercial oils:87 105 130 160 180 195 135 145 213 105 145151 152 136 87 99 92 119 129(a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places.)s^2 = . min^2s = . min(b) If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression. (Round your answer to three decimal places.)s^2 = hr^2s = hr

Step-by-step answer

P Answered by PhD

Step-by-step explanation:

Mean = (87 + 105 + 130 + 160 + 180 + 195 + 135 + 145 + 213 + 105 + 145 + 151 152 + 136 + 87 + 99 + 92 + 119 + 129)/19 = 129

Variance = (summation(x - mean)²/n

Standard deviation = √(summation(x - mean)²/n

n = 19

Variance = [(87 - 129)^2 + (105 - 129)^2 + (130 - 129)^2+ (160 - 129)^2 + (180 - 129)^2 + (195 - 129)^2 + (135 - 129)^2 + (145 - 129)^2 + (213 - 129)^2 + (105 - 129)^2 + (145 - 129)^2 + (151 - 129)^2 + (152 - 129)^2 + (136 - 129)^2 + (87 - 129)^2 + (99 - 129)^2 + (92 - 129)^2 + (119 - 129)^2 + (129 - 129)^2]/19 = 23634/19 1243.895 min

Standard deviation = √1243.895 = 35.269 min

60 minutes = 1 hour

Converting the variance to hours,

Each division would have been divided by 60². 60² can be factorized out

Variance = 23634/60² = 6.565 hours

Converting the standard deviation to hours, it becomes

√6.565 = 2.562 hours

Mathematics

An article reported the following data on oxidation-induction time (min) for various commercial oils:87 105 130 160 180 195 135 145 213 105 145151 152 136 87 99 92 119 129(a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places.)s^2 = . min^2s = . min(b) If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression. (Round your answer to three decimal places.)s^2 = hr^2s = hr

Step-by-step answer

P Answered by PhD

Step-by-step explanation:

Mean = (87 + 105 + 130 + 160 + 180 + 195 + 135 + 145 + 213 + 105 + 145 + 151 152 + 136 + 87 + 99 + 92 + 119 + 129)/19 = 129

Variance = (summation(x - mean)²/n

Standard deviation = √(summation(x - mean)²/n

n = 19

Variance = [(87 - 129)^2 + (105 - 129)^2 + (130 - 129)^2+ (160 - 129)^2 + (180 - 129)^2 + (195 - 129)^2 + (135 - 129)^2 + (145 - 129)^2 + (213 - 129)^2 + (105 - 129)^2 + (145 - 129)^2 + (151 - 129)^2 + (152 - 129)^2 + (136 - 129)^2 + (87 - 129)^2 + (99 - 129)^2 + (92 - 129)^2 + (119 - 129)^2 + (129 - 129)^2]/19 = 23634/19 1243.895 min

Standard deviation = √1243.895 = 35.269 min

60 minutes = 1 hour

Converting the variance to hours,

Each division would have been divided by 60². 60² can be factorized out

Variance = 23634/60² = 6.565 hours

Converting the standard deviation to hours, it becomes

√6.565 = 2.562 hours

Mathematics

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 120 units of the leading product provides a mean battery life of 5 hours and 40 minutes with a standard deviation of 30 minutes. A similar analysis of 100 units of the new product results in a mean battery life of 8 hours and 5 minutes

Step-by-step answer

P Answered by Master

a) Null hypothesis: $\mu_{1} - \mu_{2} \leq 120$

Alternative hypothesis: $\mu_{1} - \mu_{2}120$

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689$

c) Critical value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=100+120-2=218$

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

$t_{crit}= 1.652$

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

Step-by-step explanation:

Data given and notation

$\bar X_{1}= 8*60 +5 =485$ represent the mean for the new sample

$\bar X_{2}=5*60+40=340$ represent the mean for the old sample

$s_{1}=55$ represent the sample standard deviation for the new sample

$s_{2}=30$ represent the sample standard deviation for the old sample

$n_{1}=100$ sample size selected for the new sample

$n_{2}=120$ sample size selected for the old sample

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the new product has a battery life more than two hours longer than the leading product., the system of hypothesis would be:

Null hypothesis: $\mu_{1} - \mu_{2} \leq 120$

Alternative hypothesis: $\mu_{1} - \mu_{2}120$

If we analyze the size for the samples both are higher than 30 but we don't know the population deviations so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})- \Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689$

c) Critical value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=100+120-2=218$

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

$t_{crit}= 1.652$

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

Mathematics

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 120 units of the leading product provides a mean battery life of 5 hours and 40 minutes with a standard deviation of 30 minutes. A similar analysis of 100 units of the new product results in a mean battery life of 8 hours and 5 minutes

Step-by-step answer

P Answered by Specialist

a) Null hypothesis: $\mu_{1} - \mu_{2} \leq 120$

Alternative hypothesis: $\mu_{1} - \mu_{2}120$

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689$

c) Critical value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=100+120-2=218$

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

$t_{crit}= 1.652$

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

Step-by-step explanation:

Data given and notation

$\bar X_{1}= 8*60 +5 =485$ represent the mean for the new sample

$\bar X_{2}=5*60+40=340$ represent the mean for the old sample

$s_{1}=55$ represent the sample standard deviation for the new sample

$s_{2}=30$ represent the sample standard deviation for the old sample

$n_{1}=100$ sample size selected for the new sample

$n_{2}=120$ sample size selected for the old sample

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the new product has a battery life more than two hours longer than the leading product., the system of hypothesis would be:

Null hypothesis: $\mu_{1} - \mu_{2} \leq 120$

Alternative hypothesis: $\mu_{1} - \mu_{2}120$

If we analyze the size for the samples both are higher than 30 but we don't know the population deviations so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})- \Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

b) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689$

c) Critical value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=100+120-2=218$

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

$t_{crit}= 1.652$

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

Mathematics

A lumber company is making boards that are 2920.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A sample of 12 is made, and it is found that they have a mean of 2922.7 millimeters with a variance of 121.00. A level of significance of 0.05 will be used to determine if the boards are either too long or too short. Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places.

Step-by-step answer

P Answered by PhD

The value of the test statistic is of 0.22.

Step-by-step explanation:

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the expected mean, $\sigma$ is the standard deviation(square root of the variance) and n is the size of the sample.

A lumber company is making boards that are 2920.0 millimeters tall.

This means that $\mu = 2920$ .

A sample of 12 is made, and it is found that they have a mean of 2922.7 millimeters with a variance of 121.00.

This means that $X = 2922.7, n = 12, \sigma = \sqrt{121} = 11$ . So

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$t = \frac{2922.7 - 2922}{\frac{11}{\sqrt{12}}}$

t = 0.22

The value of the test statistic is of 0.22.

Mathematics

A lumber company is making boards that are 2564.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A sample of 21 is made, and it is found that they have a mean of 2567.0 millimeters with a variance of 121.00. A level of significance of 0.1 will be used to determine if the boards are either too long or too short. Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places.

Step-by-step answer

P Answered by Master

$t=\frac{2567-2564}{\frac{11}{\sqrt{21}}}=1.250$

The degrees of freedom are given by:

df=n-1=21-1=20

the p value for this case would be given by:

$p_v =2*P(t_{(20)}1.250)=0.2113$

For this case we see that the p value is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 2564 mm

Step-by-step explanation:

Information given

$\bar X=2567$ represent the mean height for the sample

$s=\sqrt{121}= 11$ represent the sample standard deviation

n=21 sample size

$\mu_o =2564$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to check if the true mean is equal to 2564 mm, the system of hypothesis would be:

Null hypothesis: $\mu = 2564$

Alternative hypothesis: $\mu \neq 2564$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing we got:

$t=\frac{2567-2564}{\frac{11}{\sqrt{21}}}=1.250$

The degrees of freedom are given by:

df=n-1=21-1=20

the p value for this case would be given by:

$p_v =2*P(t_{(20)}1.250)=0.2113$

For this case we see that the p value is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 2564 mm

Mathematics

Our environment is very sensitive to the amount of ozone in the upper atmosphere. the level of ozone normally found is 6.5 parts/million (ppm). a researcher believes that the current ozone level is not at a normal level. the mean of 27 samples is 6.9 ppm with a variance of 1.21. assume the population is normally distributed. a level of significance of 0.05 will be used. find the value of the test statistic. round your answer to three decimal places.

Step-by-step answer

P Answered by PhD

t = 1.890

Step-by-step explanation:

Let $\mu$ be the population mean amount of ozone in the upper atmosphere.