Question

In a recent Super Bowl, a TV network predicted that 53 % of the audience would express an interest in seeing one of its forthcoming television shows. The network ran commercials for these shows during the Super Bowl. The day after the Super Bowl, and Advertising Group sampled 79 people who saw the commercials and found that 41 of them said they would watch one of the television shows.Suppose you have the following null and alternative hypotheses for a test you are running:H0:p=0.53H0:p=0.53Ha:p≠0.53Ha:p≠0.53Calculate the test statistic, rounded to 3 decimal places

Answer 1

Answer: -0.196

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Explanation:

We're conducting a one proportion Z test.

The hypothesized population proportion is p = 0.53, which is not to be confused with the p-value (unfortunately statistics textbooks seem to overuse the letter 'p'). Luckily this problem is not asking for the p-value.

The sample population proportion is

phat = x/n = 41/79 = 0.518987 approximately

The standard error (SE) is

SE = sqrt(p*(1-p)/n)

SE = sqrt(0.53*(1-0.53)/79)

SE = 0.056153 approximately

Making the test statistic to be

z = (phat - p)/(SE)

z = (0.518987 - 0.53)/0.056153

z = -0.19612487311452

z = -0.196

Which is approximate and rounded to 3 decimal places.