09.03.2021

A diet plan claims that women have a mean weight of 145 lbs. Assume that the weights of women have a standard deviation of 30.86 lb. A sample of 40 weights of women has a mean weight of 153.2 lb. Find the P-value and, using a 0.05 significance level, state the conclusion about the null hypothesis. 0.0465; reject the null hypothesis 0.0930; fail to reject the null hypothesis

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09.07.2023, solved by verified expert
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0.0930; fail to reject the null hypothesis

Step-by-step explanation:

Sample Statistics

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59

Population Statistics

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59

Claim

The mean weight of women is 145lbs

Counterclaim:

The mean weight of women is not equal to 145lbs

Null hypothesis

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59

Alternative hypothesis

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59

Significance Level

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59 for a 2-tailed test at a 95% confidence level

Test Statistic

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59

Corresponding p-value

A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59

The test is 2-tailed because our alternate hypothesis, A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59, already implies that the population mean, A diet plan claims that women have a mean weight, №18010674, 09.03.2021 19:59, could be either less than or greater than 145lb. Since 0.09286>0.05, we are within the 95% confidence level, so there's insufficient evidence in our sample data to suggest that we reject the null hypothesis. This means that it's 9.3% more likely that the null hypothesis is true than the alternate hypothesis is. Thus, the sample data do not differ significantly from the expected mean of 145lb.

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Mathematics
Step-by-step answer
P Answered by Master

0.0930; fail to reject the null hypothesis

Step-by-step explanation:

Sample Statistics

\bar{x}=153.2

n=40

Population Statistics

\sigma=30.86

\mu=145

Claim

The mean weight of women is 145lbs

Counterclaim:

The mean weight of women is not equal to 145lbs

Null hypothesis

H_0:\mu=145

Alternative hypothesis

H_1:\mu\neq145

Significance Level

\alpha=0.05 for a 2-tailed test at a 95% confidence level

Test Statistic

Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{153.2-145}{\frac{30.86}{\sqrt{40}}}\approx1.6805

Corresponding p-value

p=2[$normalcdf(1.6805,\infty,0,1)]\approx2(0.04643)\approx0.09286\approx0.0930

The test is 2-tailed because our alternate hypothesis, H_1, already implies that the population mean, \mu, could be either less than or greater than 145lb. Since 0.09286>0.05, we are within the 95% confidence level, so there's insufficient evidence in our sample data to suggest that we reject the null hypothesis. This means that it's 9.3% more likely that the null hypothesis is true than the alternate hypothesis is. Thus, the sample data do not differ significantly from the expected mean of 145lb.

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