Question

1. A manufacturer claims that the mean lifetime of its lithium battery is 1000 hours. A homeowner selects 40 batteries and finds the mean lifetime to be 990 hours with a standard deviation of 80 hours. Test the manufacturer's claim. Use alpha = 0.05.

a.  Calculate the test statistics. test statistics= (Round your answer to the nearest hundredth) Answer is not 0.79 or 0.80

2. A local juice manufacturer distributes juice in bottles labeled 12 ounces. A government agency thinks that the company is cheating its customers. The agency selects 50 of these bottles, measures their contents, and obtains a sample mean of 11.6 ounces with a standard deviation of 0.70 ounce. Use a 0.01 significance level to test the agency's claim that the
company is cheating its customers.

a. Calculate the critical value. (Round your answer to the nearest hundredth)

(c) Calculate the test statistics. (Round your answer to the nearest hundredth)

3. A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 11.1 minutes with a standard deviation of 1.5 minutes. At the 0.01 significance level, test the claim that the mean waiting time is less than 10 minutes.

a. Calculate the test statistics. t-statistics=

b. Calculate the critical value. t-critical=

4. You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01.

      Ho:μ=86.5Ho:μ=86.5
      Ha:μ<86.5Ha:μ<86.5

You believe the population is normally distributed and you know the standard deviation is σ=14.4σ=14.4. You obtain a sample mean of M=80.7M=80.7 for a sample of size n=36n=36. NOTE: DO NOT USE T-TABLE since population is normally distributed. Use z-table to find critical value.

What is the critical value for this test? (Report answer accurate to three decimal places.)
critical value =

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

The test statistic is...

a. in the critical region

b. not in the critical region

This test statistic leads to a decision to...

a. reject the null

b. accept the null

c. fail to reject the null

As such, the final conclusion is that...

a. There is sufficient evidence to warrant rejection of the claim that the population mean is less than 86.5.

b. There is not sufficient evidence to warrant rejection of the claim that the population mean is less than 86.5.

c. The sample data support the claim that the population mean is less than 86.5.

d. There is not sufficient sample evidence to support the claim that the population mean is less than 86.5.

5. You are conducting a study to see if the probability of catching the flu this year is significantly more than 0.55. Thus you are performing a right-tailed test. Your sample data produce the test statistic z=2.363z=2.363. Find the p-value accurate to 4 decimal places.

Answer 1

1)

\mu_0 (mean) = 1000
\\ s (standard deviation) = 80
n = 40
\\ \bar{X} = 990
\\ \alpha = 0.05

The test hypothesis is


This is a two-sided test because the alternative hypothesis is formulated to detect differences from the hypothesized mean value of 30 on either side.
Now, the value of test static can be found out by following formula:

.
​​​​​​we fail to reject the null hypothesis   at