Question

Develop a hypothesis test that can be used to determine whether the mean annual consumption in Webster City is higher than the national mean. H0: µ Ha: µ What is a point estimate of the difference between mean annual consumption in Webster City and the national mean? 1.23 (2 decimals) At ???? = 0.05, test for a significant difference by completing the following. Calculate the value of the test statistic (2 decimals). .95 The p-value is .1700 (4 decimals) Reject the null hypothesis? What is your conclusion?

Answer 1

Answer:

As the P-value (0.027) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the mean annual consumption in Webster City is higher than the national mean.

Step-by-step explanation:

The question is incomplete:

Annual per capita consumption of milk is 21.6 gallons (Statistical Abstract of the United  States: 2006). Being from the Midwest, you believe milk consumption is higher there and  wish to support your opinion.

A sample of 16 individuals from the midwestern town of  Webster City showed a sample mean annual consumption of 24.1 gallons with a standard  deviation of s = 4.8.

a. Develop a hypothesis test that can be used to determine whether the mean annual  consumption in Webster City is higher than the national mean.

b. At α= .05, test for a significant difference. What is your conclusion?

a) This is a hypothesis test for the population mean.

The claim is that the mean annual consumption in Webster City is higher than the national mean.

Then, the null and alternative hypothesis are:

$H_0: \mu=21.6\\\\H_a:\mu> 21.6$

The significance level is 0.05.

The sample has a size n=16.

The sample mean is M=24.1.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=4.8.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{4.8}{\sqrt{16}}=1.2$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{24.1-21.6}{1.2}=\dfrac{2.5}{1.2}=2.083$

The degrees of freedom for this sample size are:

$df=n-1=16-1=15$

This test is a right-tailed test, with 15 degrees of freedom and t=2.083, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>2.083)=0.027$

c) As the P-value (0.027) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the mean annual consumption in Webster City is higher than the national mean.