In an article in the Journal of Advertising , Weinberger and Spotts compare the

Question

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 140 use humor, while a random sample of 500 television ads in the United States reveals that 128 use humor.

(a) Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.

H0: p1 ? p2 (Click to select)= 0 versus Ha: p1 ? p2 (Click to select)= 0.

(b) Test the hypotheses you set up in part a by using critical values and by setting ? equal to .10, .05, .01, and .001. How much evidence is there that the proportions of U.K. and U.S. ads using humor are different? (Round the proportion values to 3 decimal places. Round your answer to 2 decimal places.)

  z           

(Click to select)Do not RejectReject H0 at each value of ?; (Click to select)strongsomenoneextremely strongvery strong evidence.

(c) Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K. and U.S. ads using humor is more than .05 (five percentage points). Test these hypotheses by using a p-value and by setting a equal to .10, .05, .01, and .001. How much evidence is there that the difference between the proportions exceeds .05? (Round the proportion values to 3 decimal places. Round your z value to 2 decimal places and p-value to 4 decimal places.)

(Click to select)Do not RejectReject H0 at each value of ? = .10 and ? = .05; (Click to select)very strongextremely strongsomestrongnone evidence.

(d) Calculate a 95 percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be 95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor? (Round the proportion values to 3 decimal places. Round your answers to 4 decimal places.)

95% of Confidence Interval                      [ , ]

(Click to select)NoYes the entire interval is above zero.

Explanation / Answer

a. Since, no direction of test is specified, this is a two-tailed test.

H0:p1-p2=0 (there is no difference in proportion of UK ads using humor and proportion of US ads using humor)

H1:p1-p2=/=0 (there is difference in prportion of UK ads using humor and proportion of US ads using humor)

group1 refer to ads using humor in UK and group2 refer to ads using humor in US.

b. For large sample size, independent random samples and normal sampling distribution use Z distribution to establish the critical region. Critical values for two tailed test at alpha=0.10 is +-1.65, for alpha=0.05 is +-1.96, for alpha=0.01 is +-2.58 and for alpha=0.001 is +-3.32.

The Z test statistic is as follows:

Pooled sample proportion:phatp=(x1+x2)/(n1+n2), where, x refer to number of successes and n is number of trials.

=(140+128)/(400+500)

=0.2978

z=(p1hat-p2hat)/sqrt{phatp(1-phatp)}sqrt(1/n1+1/n2)

=(140/400-128/500)/sqrt{0.2978(1-0.2978)}sqrt(1/400+1/500)

=(0.350-0.256)/sqrt{0.2978(1-0.2978)}sqrt(1/400+1/500)

=3.06

Per rule based on critical value, reject H0, if observed z>=critical z. Here, 3.06 is greater than 1.65, 1.96 and 2.58.

c. z=3.06, p=0.002

Per rejection rule based on p value, reject H0, if p value is less than alpha.

Therefore, reject H0 at alpha=0.10, and alpha=0.05.There is very strong evidence.

d. The 95% c.i=(p1hat-p2hat)+-zalpha/2 sqrt[p1hat(1-p1hat)/n1+p2hat(1-p2hat)/n2]

=(140/400-128/500)+-1.96sqrt[140/400(1-140/400)/400+128/500(1-128/500)/500]

=(0.0336,0.1544)

Per rejection rule based on c.i, reject H0, if confidence interval doesnot contain 0. Thus, reject H0, since the entire interval is above 0.

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