An economist wondered if people who go to the movies on weekdays go more or less often on Fridays than any other day. She figured that if it were truly random, 20% of these movie-goers would go on Fridays. She randomly sampled 50 people who go to movies on weekdays and asked them on which day they attend movies most frequently. Of those sampled, 14 indicated that they go on Fridays more often than other days.
The economist conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of weekday movie-goers who go most frequently on Fridays is different from 20%.
For this test: H0:p=0.2; Ha:p≠0.2, which is a two-tailed test.
Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether the true proportion of weekday movie-goers who go most frequently on Fridays is different from 20%. Identify the test statistic, z, and p-value from the calculator output, rounding to three decimal places.
Given that, n = 50 and x = 14
=> sample proportion = 14/50 = 0.28
The null and alternative hypotheses are,
H0 : p = 0.2
Ha : p ≠ 0.2
Using TI-83 plus calculator we get,
Test statistic = 1.4142
p-value = 0.157
Since, p-value > 0.05, we fail to reject H0.
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