Question

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 100 recent prime time airings from both networks have been considered. The first network aired an average of 110.8 commercials during prime with a standard deviation of 5.0. The second network aired 109.2 commercials with a standard deviation of 4.6. Since the sample size is quite large, assume that the population standard deviations 5.0 and 4.6 can be estimated using the sample standard deviations. At the 0.05 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, μ 1 is not equal to the average number of commercials aired during prime time by the second station μ2? Perform a two-tailed test. Then fill in the table below Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) The null hypothesis: The alternative hypothesis:H, The type of test statiCooss one (Choose one) V The value of the test statistic: Round to at least three decimal places.) two critical values at the 0.05 level of significance:and Round to at least three decimal places.) Can we support t of The claim that the mean number commercials aired network is not equal to the mean number of commercials aired d second network? during prime time by the first O Yes No uring prime time b y the

0 0
Add a comment Improve this question Transcribed image text
Answer #1

To test H_0:\mu_1 = \mu_2 against H_1:\mu_1 \neq \mu_2

Here

\bar{x_1} = 110.8, \sigma_1 = 5,n_1=100

and \bar{x_2} = 109.2, \sigma_2= 4.6,n_2=100

The type of test statistic : z statistic

The test statistic can be written as

(1 - 2) m 2 7t which under H0 follows a standard normal distribution.

We reject H0 at 5% level of significance if |z_{obs}|>z_{0.025}

Now,

The value of the test statistic z_{obs}=2.35498

Critical value \pm z_{0.025} = \pm 1.95996

Since z_{obs}=2.35498 > z_{0.025} = 1.95996 , so we reject H0 at 5% level of significance and we support the claim that the mean number of commercials aired sring prime time by the first network is not equal to the mean number of commercials aired sring prime time by the second network.

ans-> Yes

Add a comment
Know the answer?
Add Answer to:
The historical reports from two major networks showed that the mean number of commercials aired during...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Random and independent samples of 95 recent prime time airings from each of two major networks...

    Random and independent samples of 95 recent prime time airings from each of two major networks have been considered. The first network aired a mean of 110.6 commercials during prime time, with a standard deviation of 4.3 commercials. The second network aired a mean of 109.4 commercials, with a standard deviation of 4.7 commercials. As the sample sizes are quite large, the population standard deviations can be estimated using the sample standard deviations. Construct a 95% confidence interval for Hly,...

  • November 6, 2012 was election day. Many of the major television networks aired coverage of the...

    November 6, 2012 was election day. Many of the major television networks aired coverage of the incoming election results during the primetime hours. For a random sample of 25 U.S. adults, the average time spent watching election coverage was 80.44 minutes with standard deviation of 43.99 minutes. We want to do a hypothesis test to see if this is sufficient evidence to reject the null hypothesis in favor of the alternative: H0: the mean number of minutes watching election coverage...

  • A leasing firm claims that the mean number of miles driven annually, H, in its leased...

    A leasing firm claims that the mean number of miles driven annually, H, in its leased cars is less than 13420 miles. A random sample of 25 cars leased from this firm had a mean of 13149 annual miles driven. It is known that the population standard deviation of the number of miles driven in cars from this firm is 1260 miles. Assume that the population is normally distributed. Is there support for the firm's claim at the 0.05 level...

  • A leasing firm claims that the mean number of miles driven annually, , in its leased...

    A leasing firm claims that the mean number of miles driven annually, , in its leased cars is less than 13560 miles. A random sample of 100 cars leased from this firm had a mean of 13428 annual miles driven. It is known that the population standard deviation of the number of miles driven in cars from this firm is 2280 miles. Is there support for the firm's claim at the 0.05 level of significance? Perform a one-tailed test. Then...

  • The human resources department of a major corporation announced that the number of people interviewed by...

    The human resources department of a major corporation announced that the number of people interviewed by the corporation in one month has a mean of 104 and a standard deviation, o, of 15. The management of the corporation suspects that the standard deviation differs from 15. A random sample of 22 months had a mean of 99 interviews, with a standard deviation of 11. If we assume that the number of people interviewed by the corporation in one month follows...

  • Medical researchers interested in determining the relative effectiveness of two different drug treatments on people with...

    Medical researchers interested in determining the relative effectiveness of two different drug treatments on people with a chronic mental illness established two independent test groups. The first group consisted of 14 people with the illness, and the second group consisted of 13 people with the illness. The first group received treatment 1 and had a mean time until remission of 172 days, with a standard deviation of 6 days. The second group received treatment and had a mean time until...

  • Records from previous years for a casualty insurance company show that its clients average a combined...

    Records from previous years for a casualty insurance company show that its clients average a combined total of 1.7 auto accidents per day, with a standard deviation of 0.56. The actuaries of the company claim that the current standard deviation, o, of the number of accidents per day is not equal to 0.56. A random sample of 10 recent days had a mean of 2 accidents per day with a standard deviation of 0.26. If we assume that the number...

  • Assume that the two samples are independent simple random samples selected from normally distributed populations. Do...

    Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal Refer to the accompanying data set. Use a 0.05 significance level to test the claim that women and me Click the icon to view the data for diastolic blood pressure for men and women Data for Diastolic Blood Pressure of Men and Women Let , be the mean diastolic blood pressure for women and let...

  • A random selection of volunteers at a research institute have been exposed to a typical cold...

    A random selection of volunteers at a research institute have been exposed to a typical cold virus. After they started to have cold symptoms, 25 of them were given multivitamin tablets daily which contain 3 grams of vitamin C and various other vitamins and minerals. The remaining 25 volunteers were given placebo tablets. For each individual, the length of time taken to recover from cold is recorded. At the end of the experiment following data are obtained: Days to recover...

  • For the following test, assume that the two samples are independent simple random samples selected from...

    For the following test, assume that the two samples are independent simple random samples selected from normally distributed populations and that the population standard deviations are equal (1 = 02) so that the standard error of the difference between means is obtained by pooling the sample variances. A researcher was interested in comparing the amount of time spent studying baboons by women and men from a group of anthropologists. Independent simple random samples of 32 women and 30 men were...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT