Question

3. A paint manufacturing company claims that the mean drying time for its paints is not longer than 60 minutes. A random sample of 20 gallons of paints selected from the production line of this company showed that the mean drying time for this sample is 63.50 minutes with a sample standard deviation of 4 min utes. Assume that the drying times for these paints have a normal distribution. Test the claim that the mean drying time for its paints is less 60 minutes. 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. Make a decision to reject or fail to reject the H0.
e. Interpret the decision in the context of the original claim



4. The manager of a restaurant in a large city claims that waiters working in all restaurants in his city earn an average of $150 or more in tips per week. A random sample of 25 waiters selected from restaurants of this city yielded a mean of $139 in tips per week with a sample standard deviation of $28. Assume that the weekly tips for all waiters in this city have a normal distribution. Test the claim that the mean tips earned by waiters in that city is $150 or more. 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

3.

(A)Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ = 60 or mean drying time for its paints is 60

Ha:μ < 60 or mean drying time for its paints is not longer than 60

(B)

the significance level is α=0.01, and the critical value for a left-tailed test is tc​=−2.539.

(C)

Test Statistics

The t-statistic is computed as follows:

(D)

Since it is observed that t=3.913≥tc​=−2.539, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.9995, and since p=0.9995≥0.01, it is concluded that the null hypothesis is not rejected.

(E)

Hence, since the null hypothesis is accepted.

we claim that the mean drying time for its paints is 60 minutes.

4.

(A)

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 150 or  waiters working in all restaurants in his city earn an average of $150 .

Ha: μ > 150 or  waiters working in all restaurants in his city earn an average of $150 or more.

(B)

the significance level is α=0.01, and the critical value for a right-tailed test is tc​=2.492.

(C)

Test Statistics

The t-statistic is computed as follows:

(D)

Since it is observed that t=−1.964≤tc​=2.492, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.9694, and since p=0.9694≥0.01, it is concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean \muμ is greater than 150, at the 0.01 significance level.

(E)

since the null hypothesis is accepted.

hence,waiters working in all restaurants in his city earn an average of $150 .

please rate my answer and comment for doubts.