Homework SourseA husband and wife Ted and Suzanne share a digital music pla
A husband and wife, Ted and Suzanne, share a digital music player that has a feature that randomly selects which song to play. A total of 3476 songs have been loaded into the player, some by Ted and the rest by Suzanne. They are interested in determining whether they have each loaded a different proportion of songs into the player. Suppose that when the player was in the random-selection mode, 22 of the first 29 songs selected were songs loaded by Suzanne. Let p denote the proportion of songs that were loaded by Suzanne. State the null and alternative hypotheses to be tested. How strong is the evidence that Ted and Suzanne have each loaded a different proportion of songs into the player? (Round your test statistic to two decimal places and your P-value to four decimal places. Assume a 95% confidence level.)
Hypotheses:
H0: p = 0.5
Ha: p < 0.5H0: p = 0.5
Ha: p > 0.5 H0: p = 0.5
Ha: p 0.5
Conclusion:
There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.
There is not enough evidence to conclude that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.
| z | = |
| P-value | = |
Solution
Formulating the null and alternatuve hypotheses,
Ho: p = 0.5
Ha: p =/= 0.5
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As we see, the hypothesized po = 0.5
Getting the point estimate of p, p^,
p^ = x / n = 0.75862069
Getting the standard error of p^, sp,
sp = sqrt[po (1 - po)/n] = 0.092847669
Getting the z statistic,
z = (p^ - po)/sp = 2.785430073
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As this is a 2 tailed test, then, getting the p value,
p = 0.002672838
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significance level = 0.05
Comparing p and the significance value, we REJECT THE NULL HYPOTHESIS.
Conclusion:
There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.
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