Mathematics : asked on rb276
 06.04.2020

The value x to one decimal place

. 4

Faq

Mathematics
Step-by-step answer
P Answered by Master

a) 17-2.26\frac{1.9}{\sqrt{10}}=15.64  

17+2.26\frac{1.9}{\sqrt{10}}=18.36  

So on this case the 95% confidence interval would be given by (15.64;18.36)

b) 1. n=15, conf =95% \bar X= 35 s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% \bar X= 82 s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% \bar X= 0.9 s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

Step-by-step explanation:

Part a: What is the lower bound to this confidence interval? 2 cm (round to 2 decimal places) What is the upper bound to this confidence interval? cm (round to 2 decimal places)

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

We have the following data:

\bar x= 17 represent the sample mean

s = 1.9 represent the sample deviation

n =10 represent the sample size

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=10-1=9  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that t_{\alpha/2}=2.26  

Now we have everything in order to replace into formula (1):  

17-2.26\frac{1.9}{\sqrt{10}}=15.64  

17+2.26\frac{1.9}{\sqrt{10}}=18.36  

So on this case the 95% confidence interval would be given by (15.64;18.36)

Part b

1. n=15, conf =95% \bar X= 35 s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% \bar X= 82 s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% \bar X= 0.9 s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

Mathematics
Step-by-step answer
P Answered by Specialist

a) 17-2.26\frac{1.9}{\sqrt{10}}=15.64  

17+2.26\frac{1.9}{\sqrt{10}}=18.36  

So on this case the 95% confidence interval would be given by (15.64;18.36)

b) 1. n=15, conf =95% \bar X= 35 s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% \bar X= 82 s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% \bar X= 0.9 s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

Step-by-step explanation:

Part a: What is the lower bound to this confidence interval? 2 cm (round to 2 decimal places) What is the upper bound to this confidence interval? cm (round to 2 decimal places)

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

We have the following data:

\bar x= 17 represent the sample mean

s = 1.9 represent the sample deviation

n =10 represent the sample size

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=10-1=9  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that t_{\alpha/2}=2.26  

Now we have everything in order to replace into formula (1):  

17-2.26\frac{1.9}{\sqrt{10}}=15.64  

17+2.26\frac{1.9}{\sqrt{10}}=18.36  

So on this case the 95% confidence interval would be given by (15.64;18.36)

Part b

1. n=15, conf =95% \bar X= 35 s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% \bar X= 82 s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% \bar X= 0.9 s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

Mathematics
Step-by-step answer
P Answered by Master

a) Null hypothesis:\mu \leq 76  

Alternative hypothesis:\mu  76

b) t=\frac{77.5-76}{\frac{3.3}{\sqrt{29}}}=2.45  

c) p_v =P(t_{(28)}2.45)=0.0104  

d)  Fail to reject H0. There is not enough evidence to support the claim.

Step-by-step explanation:

Data given and notation  

\bar X=77.5 represent the sample mean  

s=3.3 represent the sample standard deviation  

n=29 sample size  

\mu_o =76 represent the value that we want to test  

\alpha=0.01 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 population mean is higher than 76, the system of hypothesis are :  

Null hypothesis:\mu \leq 76  

Alternative hypothesis:\mu  76  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

b) Calculate the statistic  

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

t=\frac{77.5-76}{\frac{3.3}{\sqrt{29}}}=2.45  

c) P-value  

We need to calculate the degrees of freedom first given by:

df=n-1=29-1=28

Since is a one-side right tailed test the p value would given by:  

p_v =P(t_{(28)}2.45)=0.0104  

We can use the following code in excel to find the "=1-T.DIST(2.45,28,TRUE)"

d) Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

The best option on this case:

Fail to reject H0. There is not enough evidence to support the claim.

Mathematics
Step-by-step answer
P Answered by Master

a) Null hypothesis:\mu \leq 76  

Alternative hypothesis:\mu  76

b) t=\frac{77.5-76}{\frac{3.3}{\sqrt{29}}}=2.45  

c) p_v =P(t_{(28)}2.45)=0.0104  

d)  Fail to reject H0. There is not enough evidence to support the claim.

Step-by-step explanation:

Data given and notation  

\bar X=77.5 represent the sample mean  

s=3.3 represent the sample standard deviation  

n=29 sample size  

\mu_o =76 represent the value that we want to test  

\alpha=0.01 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 population mean is higher than 76, the system of hypothesis are :  

Null hypothesis:\mu \leq 76  

Alternative hypothesis:\mu  76  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

b) Calculate the statistic  

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

t=\frac{77.5-76}{\frac{3.3}{\sqrt{29}}}=2.45  

c) P-value  

We need to calculate the degrees of freedom first given by:

df=n-1=29-1=28

Since is a one-side right tailed test the p value would given by:  

p_v =P(t_{(28)}2.45)=0.0104  

We can use the following code in excel to find the "=1-T.DIST(2.45,28,TRUE)"

d) Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

The best option on this case:

Fail to reject H0. There is not enough evidence to support the claim.

Mathematics
Step-by-step answer
P Answered by PhD

(A) Null Hypothesis, H_0 : \mu_1 \geq \mu_2    

     Alternate Hypothesis, H_A : \mu_1  

(B) The value of t-test statistics is -18.48.

(C) The P-value is Less than 0.005%.

(D) Reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda.

Step-by-step explanation:

We are given that the Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right;

Diet Regular

μ μ1 μ2

n 20 20

x 0.78062lb 0.81645 lb

s 0.00444 lb 0.00745 lb

Let \mu_1 = mean weight of contents of cans of diet soda.

\mu_2 = mean weight of contents of cans of regular soda.

So, Null Hypothesis, H_0 : \mu_1 \geq \mu_2      {means that the contents of cans of diet soda have weights with a mean that is more than or equal to the mean for the regular soda}

Alternate Hypothesis, H_A : \mu_1     {means that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

                    T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean weight of cans of diet soda = 0.78062 lb

\bar X_2 = sample mean weight of cans of regular soda = 0.81645 lb

s_1 = sample standard deviation of cans of diet soda = 0.00444 lb

s_2 = sample standard deviation of cans of regular soda = 0.00745 lb

n_1 = sample of cans of diet soda = 20

n_2 = sample of cans of diet soda = 20

Also,  s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} } = \sqrt{\frac{(20-1)\times 0.00444^{2}+ (20-1)\times 0.00745^{2}}{20+20-2} } = 0.00613

So, the test statistics =  \frac{(0.78062-0.81645)-(0)}{0.00613 \times \sqrt{\frac{1}{20}+\frac{1}{20} } }  ~  t_3_8

                                    =  -18.48

The value of t-test statistics is -18.48.

Also, the P-value of the test statistics is given by;

              P-value = P( t_3_8 < -18.48) = Less than 0.005%

Now, at a 0.01 level of significance, the t table gives a critical value of -2.429 at 38 degrees of freedom for the left-tailed test.

Since the value of our test statistics is less than the critical value of t as -18.48 < -2.429, so we have sufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda.

Mathematics
Step-by-step answer
P Answered by PhD

(A) Null Hypothesis, H_0 : \mu_1 \geq \mu_2    

     Alternate Hypothesis, H_A : \mu_1  

(B) The value of t-test statistics is -18.48.

(C) The P-value is Less than 0.005%.

(D) Reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda.

Step-by-step explanation:

We are given that the Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right;

Diet Regular

μ μ1 μ2

n 20 20

x 0.78062lb 0.81645 lb

s 0.00444 lb 0.00745 lb

Let \mu_1 = mean weight of contents of cans of diet soda.

\mu_2 = mean weight of contents of cans of regular soda.

So, Null Hypothesis, H_0 : \mu_1 \geq \mu_2      {means that the contents of cans of diet soda have weights with a mean that is more than or equal to the mean for the regular soda}

Alternate Hypothesis, H_A : \mu_1     {means that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

                    T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean weight of cans of diet soda = 0.78062 lb

\bar X_2 = sample mean weight of cans of regular soda = 0.81645 lb

s_1 = sample standard deviation of cans of diet soda = 0.00444 lb

s_2 = sample standard deviation of cans of regular soda = 0.00745 lb

n_1 = sample of cans of diet soda = 20

n_2 = sample of cans of diet soda = 20

Also,  s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} } = \sqrt{\frac{(20-1)\times 0.00444^{2}+ (20-1)\times 0.00745^{2}}{20+20-2} } = 0.00613

So, the test statistics =  \frac{(0.78062-0.81645)-(0)}{0.00613 \times \sqrt{\frac{1}{20}+\frac{1}{20} } }  ~  t_3_8

                                    =  -18.48

The value of t-test statistics is -18.48.

Also, the P-value of the test statistics is given by;

              P-value = P( t_3_8 < -18.48) = Less than 0.005%

Now, at a 0.01 level of significance, the t table gives a critical value of -2.429 at 38 degrees of freedom for the left-tailed test.

Since the value of our test statistics is less than the critical value of t as -18.48 < -2.429, so we have sufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda.

Mathematics
Step-by-step answer
P Answered by Master

Given:

Sham: n= 20,     x=0.44,   s=1.24,

Magnet:n= 20,  x =0.49,   s= 0.95

For Sham:

Sample size, n = 20

Sample mean = 0.44

Standard deviation = 1.24

For Magnet:

Sample size = 20

Sample mean = 0.49

Standard deviation = 0.95

The null and alternative hypotheses:

H0: s1²=s2²

H1: s1² ≠ s2²

a) To find the test statistics, use the formula:

\frac{s1^2}{s2^2}

\frac{1.24^2}{0.95^2} = \frac{1.5376}{0.9025} = 1.7037

Test statistics = 1.7037

b) P-value:

Sham: degrees of freedom = n - 1 = 20 - 1 = 19

Magnet: degrees of freedom = n - 1 = 20 - 1 = 19

The critical values:

[Za/2, df1, df2)], [(1 - Za/2), df1, df2]

f[0.05/2, 19, 19], f[(1 - 0.05/2), 19, 19]

f[0.025, 19, 19], f[0.975, 19, 19]

(2.526, 0.3958)

The rejection region:

Reject H0, if  F < 0.3958 or if F > 2.526

c) Conclusion:

Since the critical values of test statistic is between (0.3958 < 1.7037 < 2.526), we fail to reject null hypothesis H0.

There is insufficient evidence to to support the claim that those given a sham treatment have reductions that vary more than those treated with magnets

Mathematics
Step-by-step answer
P Answered by Master

Given:

Sham: n= 20,     x=0.44,   s=1.24,

Magnet:n= 20,  x =0.49,   s= 0.95

For Sham:

Sample size, n = 20

Sample mean = 0.44

Standard deviation = 1.24

For Magnet:

Sample size = 20

Sample mean = 0.49

Standard deviation = 0.95

The null and alternative hypotheses:

H0: s1²=s2²

H1: s1² ≠ s2²

a) To find the test statistics, use the formula:

\frac{s1^2}{s2^2}

\frac{1.24^2}{0.95^2} = \frac{1.5376}{0.9025} = 1.7037

Test statistics = 1.7037

b) P-value:

Sham: degrees of freedom = n - 1 = 20 - 1 = 19

Magnet: degrees of freedom = n - 1 = 20 - 1 = 19

The critical values:

[Za/2, df1, df2)], [(1 - Za/2), df1, df2]

f[0.05/2, 19, 19], f[(1 - 0.05/2), 19, 19]

f[0.025, 19, 19], f[0.975, 19, 19]

(2.526, 0.3958)

The rejection region:

Reject H0, if  F < 0.3958 or if F > 2.526

c) Conclusion:

Since the critical values of test statistic is between (0.3958 < 1.7037 < 2.526), we fail to reject null hypothesis H0.

There is insufficient evidence to to support the claim that those given a sham treatment have reductions that vary more than those treated with magnets

Mathematics
Step-by-step answer
P Answered by Specialist

a) Alternative hypothesis, H_{a} : \mu_{1} \neq  \mu_{2}

b) z = -2.91, Pvalue = 0.002

c)Option C. Reject H_0. There is enough evidence at the 1% level of significance to reject the claim.

Step-by-step explanation:

a) Null hypothesis, H_{0} : \mu_{1} = \mu_{2}

Alternative hypothesis, H_{a} : \mu_{1} \neq  \mu_{2}

b) Standardized test statistic

Level of significance, \alpha = 0.01

\sigma_{1}  = 3.2

\sigma_{2} = 1.7

X_{1} = 16\\X_{2} = 18\\n_{1} = 30\\n_{2} = 27

z = \frac{\mu_{1}- \mu_{2}  }{\sqrt{\frac{\sigma_{1} ^{2} }{n_{1} } + \frac{\sigma_{2} ^{2} }{n_{2} } }}

z = \frac{16-18  }{\sqrt{\frac{3.2^{2} }{30 } + \frac{1.7 ^{2} }{27 } }}

z = -2.91

Checking the p-value that corresponds to z = -2.906

P-value = 0.002

c) What is the proper decision

P-value  = 0.002

Level of significance, α = 0.01

0.002 < 0.01

Since Pvalue < α, the null hypothesis H₀ should be rejected.

Option C is the correct option.

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