A MP3 Manufacturer claims that 65% of teenagers have their own MP3 player. A researcher wishes to test the claim and selects a random sample of 80 teenagers. She finds that 57 have their MP3 player. At a .05 significance level, should the claim be rejected? Please show work.

Null hypothesis: H0 = 0.65

Alternative hypothesis: Ha ≠ 0.65

z = 1.172

P value = P(Z≠1.172) = 0.24

Decision we fail to reject the null hypothesis. That is, there is convincing evidence enough to reject the Null hypothesis.

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 95% confidence interval) reject Null hypothesis

Z score < Z(at 95% confidence interval) accept Null hypothesis

Step-by-step explanation:

Given;

n=80 represent the random sample taken

Null hypothesis: H0 = 0.65

Alternative hypothesis: Ha ≠ 0.65

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 80

po = Null hypothesized value = 0.65

p^ = Observed proportion = 57/80 = 0.7125

Substituting the values we have

z = (0.7125-0.65)/√(0.65(1-0.65)/80)

z = 1.17201807734

z = 1.172

To determine the p value (test statistic) at 0.05 significance level, using a two tailed hypothesis.

P value = P(Z≠1.172) = 0.241197 = 0.24

Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = 1.172) which falls within the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.24 which is higher than 0.05. Then we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that at 5% significance level the null hypothesis is valid.

step-by-step explanation:

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step-by-step explanation: