Question

Hypothesis Testing

1. According to the U.S. Postal Service, the mean weight of mail received by Americans in 2017 through the Postal Service was 57.2 pounds. One hundred randomly selected Americans were asked to keep all their mail for last year. It was found that they received an average of 56.1 pounds of mail last year. Suppose that the population standard deviation is 9.2 pounds. Test the claim that the mean weight of mail received by Americans in 2017 through the Postal Service is different from 57.2 pounds. Use α=0.10. Answer the following questions.

a. Identify the claim and state the H0 and H1.  
b. Find the critical value.
c. Calculate the test statistic.
d. e. Make a decision to reject or fail to reject the HInterpret the decision in the context of the original claim.0.   



2. According to Moebs Services Inc., an individual checking account at major U.S. banks costs these banks between $350 and $450 per year. Suppose that the mean cost of individual checking accounts at major U.S. banks was $400 for the year 2011. A bank consultant wants to determine whether the current mean cost of such checking accounts at major U.S. banks is more than $400 a year. A recent random sample of 150 such checking accounts taken from major U.S. banks produced a mean annual cost to them of $390. Assume that the population standard deviation of annual costs to major banks of all such checking accounts is $50. Test the claim that the mean cost of checking account is more than $400. Use α=0.01.

a. Identify the claim and state the H0 and H1.  
b. Find the critical value.
c. Calculate the test statistic.
d. e. Make a decision to reject or fail to reject the HInterpret the decision in the context of the original claim.0.

Answer 1

1). We are given that

100, T = 56.1, 0 = 9.2 and yo = 57.2

a. The claim is that the mean weight of mail received by Americans in 2017 through the Postal Service is different from 57.2 pounds.

$H_{0}:\mu=57.2$

H, :μ# 57.2

b. An Z-test will be used since, we know the population sd.

It is a two sided alternative and the critical value for the test is Z=1.6449

c. The test statistic is $Z=\frac{\bar{x}-\mu_{0}}{\sigma/\sqrt{n}}$

Z 56.1 – 57.2 9.2/V100

$Z=\frac{-1.1}{9.2/10}$

$Z=-1.1957$

d. Decision: Since $\left |Z \right |=1.1957<Z_{critical}$ , we fail to reject the null hypothesis.

e. Hence, we conclude that there is not enough evidence to claim that the population mean weight is different than 57.2, at the 0 .1 significance level.

2). We are given that $n=150, \bar{x}=390,\sigma=50 \: \: and\: \: \mu_{0}=400$

a. The claim here is  "the mean cost of checking account is more than $400".

$H_{0}:\mu=400$

$H_{1}:\mu\ne400$

b. An Z-test will be used since, we know the population sd.

It is an one sided alternative and the critical value for the test at 0.01 level is is Z=2.3263

c. The test statistic is $Z=\frac{\bar{x}-\mu_{0}}{\sigma/\sqrt{n}}$

$Z=\frac{390-400}{50/\sqrt{150}}$

$Z=\frac{-10}{50/12.2474}$

$Z=-2.4495$

d. Since the calculated value of Z <the critical value, we fail to reject the null hypothesis.

e. Hence, we conclude that there is not enough evidence to claim that the population mean cost of checking is greater than $400, at the 0.01 significance level.