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.

i get you, what you should do is think of the positive things that would come out of it. it will make yours parents proud and raise your grade. you should always try and if you can't focus tell the teacher.

9987

step-by-step explanation: