Question

A market research analyst believes that the average weekly household expenditure on groceries in Pearland is $125. Assume that the weekly household expenditures are normally distributed. In setting up the hypotheses to test the analyst’s belief, she obtained a test statistic value of z = 1.45 Calculate the p-value based on normal distribution.

A. 0.4265 B. 0.1470 C. 0.0287 D. 0.0735 E. 0.0574

Based on the results of the hypothesis test at the 10% level of significance, which of the following is a correct conclusion?

A. There is sufficient evidence to conclude that the average weekly household expenditure on groceries in Pearland is $125.

B.There is not sufficient evidence to conclude that the average weekly household expenditure on groceries in Pearland is significantly different from $125.

C. There is sufficient evidence to conclude that the average weekly household expenditure on groceries in Pearland is significantly different from $125.

Answer 1

Since is a two-sided test the p value would be:

`p_v =2*P(z>1.45)=0.147`

B. 0.1470

B.There is not sufficient evidence to conclude that the average weekly household expenditure on groceries in Pearland is significantly different from $125.

Explanation:

Notation

`\bar X` represent the sample mean

`\sigma` represent the population standard deviation for the sample

`n` sample size

`\mu_o =125` represent the value that we want to test

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

z would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 125 or not, the system of hypothesis would be:

Null hypothesis:
`\mu =125`

Alternative hypothesis:
`\mu \\eq 125`

z-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".

Calculate the statistic

The statistic for this case is
`z_(calc)= 1.45`

P-value

Since is a two-sided test the p value would be:

`p_v =2*P(z>1.45)=0.147`

B. 0.1470

Conclusion

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis. And the best conclusion would be:

B.There is not sufficient evidence to conclude that the average weekly household expenditure on groceries in Pearland is significantly different from $125.