For serious students: Testing of hypothesis

The historical reports from two major networks showed that the mean number of commercials aired during prime time was equal for both networks last year. In order to find out whether they still air the same number of commercials on average or not, random and independent samples of recent prime time airings from both networks have been considered. The first network aired an average of commercials during prime with a standard deviation of . The second network aired commercials with a standard deviation of . Since the sample size is quite large, assume that the population standard deviations and can be estimated using the sample standard deviations. At the level of significance, is there sufficient evidence to support the claim that the average number of commercials aired during prime time by the first station, is not equal to the average number of commercials aired during prime time by the second station ? Perform a two-tailed test. Then fill in the table below.

What's the null hypothesis?
What's the alternative hypothesis?
What type of test statistic? Choose one: ___Z ___ t ___ Chi Square ___F
What is the value of the test statistic (round to at least 3 decimal places)?
What are the two critical values at the 0.10 level of significance? (round to at least 3 decimal places)?
Can we support the claim that the mean number of commercials aired during prime time by the first network is not equal to the mean number of commercials aired during prime time by the second network? YES or NO
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The university data center has two main computers. The center wants to examine whether computer 1 is receiving tasks which require comparable processing time to those of computer 2. A random sample of processing times from computer 1 showed a mean of seconds with a standard deviation of seconds, while a random sample of processing times from computer 2 (chosen independently of those for computer 1) showed a mean of seconds with a standard deviation of seconds. Assume that the populations of processing times are normally distributed for each of the two computers, and that the variances are equal. Can we conclude, at the level of significance, that the mean processing time of computer 1, , is greater than the mean processing time of computer 2, ?
Perform a one-tailed test

What's the null hypothesis?
What's the alternative hypothesis?
What type of test statistic? Choose one: ___Z ___ t ___ Chi Square ___F
What is the value of the test statistic (round to at least 3 decimal places)?
What is p-value? (round to at least 3 decimal places)?
Can we conclude that the mean processing time of computer 1 is greater than the mean processing time of computer 2? Yes or No
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The university data center has two main computers. The center wants to examine whether computer 1 is receiving tasks which require comparable processing time to those of computer 2. A random sample of processing times from computer 1 showed a mean of seconds with a standard deviation of seconds, while a random sample of processing times from computer 2 (chosen independently of those for computer 1) showed a mean of seconds with a standard deviation of seconds. Assume that the populations of processing times are normally distributed for each of the two computers, and that the variances are equal. Can we conclude, at the level of significance, that the mean processing time of computer 1, , is less than the mean processing time of computer 2, ?
Perform a one-tailed test. Then fill in the table below.
What's the null hypothesis?
What's the alternative hypothesis?
What type of test statistic? Choose one: ___Z ___ t ___ Chi Square ___F
What is the value of the test statistic (round to at least 3 decimal places)?
What is the critical value at the 0.05 level of significance? (round to at least 3 decimal places)?
Can we conclude that the mean processing time of computer 1 is greater than the mean processing time of computer 2? Yes or No

See attached file for full problem description.

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Solution Summary

To find the p-value of Z, t test and determine if the mean is significantly different.