Question

Executives at Sale Mart Supermarket claim that a typical family of four spends $200 weekly on routine, non-holiday, grocery purchases. According to published industry standards, the population standard deviation is $25. Stanley, another stats intern at corporate headquarters, wonders if that original claim by the executives is too high. As a project, she collects from store sales receipts a simple random sample (SRS) of size 100. The sample mean for the weekly grocery purchases for a family of four is $197. He is defining as rare, or unusually low, any sample mean that is in the bottom 10% of all possible sample means; hence, he is testing at the 10% level of significance.


What is the critical value, rounded to three decimal places and expressed without insignificant digits?

Answer 1

Answer:

the ans is 0.2259. is the correct answer

please mark me brainlist one point needed for going other level

Answer 2

The value of the test statistic

z=(205-200)/25/✓64=1.6

The hypotheses

H0:

a typical family of four spends $200 weekly on routine, non-holiday, grocery purchases. (M=200)

H1:

a typical family of four spends more than $200 weekly on routine, non-holiday, grocery purchases. (M>200)

• the critical value

z0.05=1.645

• the test statistic compares against the critical value in the right tail under the standard normal graph two-tailed confidence level corresponds to a one-sided significance level of 5%

90%(because 5%+5%=10% and 100%-10%=90%)

• The lower bound of the confidence interval

205-(1.645*25/✓64)=199.86

• The upper bound of the confidence interval

205+(1.645*25/✓64)=210.14

• the technical conclusion to this statistical inference

Since

If the test statistic is not greater than the critical value then we fail to reject the null hypothesis.

• the contextual conclusion to this statistical inference

At a 5% level of significance we conclude there I,s sufficient evidence to support the claim of Executives at Sale Mart Supermarket

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