Question 1042151: 600 people randomly polled from acityshowed that 330 supported gun control.Test the claim that the majority (more than 50%) in the city supported gun control. Use a significance level of 0.05.
(a) What’s the sample proportion, ?
(b) State your hypotheses. H0:
Ha:
(c) Check conditions required for computations.
(d) Compute the observed value of the test statistic.z =
(e) Find the p-value.p-value =
(f) Do you reject or not reject H0?
(g) Interpret your decision in part (f).
(h) Find a 90% confidence interval for the percentage of people in the city who supported gun control.
(i) Interpret your result in part (h).
(j) Do the conclusions in parts (g) and (i) agree with each other?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 600 people randomly polled from acityshowed that 330 supported gun control.Test the claim that the majority (more than 50%) in the city supported gun control. Use a significance level of 0.05.
(a) What’s the sample proportion, 330/600 = 0.55
(b) State your hypotheses.
H0: p <= 0.5
Ha: p > 0.5 (claim)
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(c) Check conditions required for computations.
(d) Compute the observed value of the test statistic.
z(0.55) = (0.55-0.50)/sqrt(0.5*0.5/600) = 2.449
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(e) Find the p-value.p-value = p(z > 2.449) = normalcdf(2.449,100) = 0.007
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(f) Do you reject or not reject H0? reject
(g) Interpret your decision in part (f). Since the p-value is less
than 5%, reject Ho.
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(h) Find a 90% confidence interval for the percentage of people in the city who supported gun control.
ME = 1.645*sqrt(0.5*0.5/600) = 0.0336
90%CI:: 0.55-0.0336 < p < 0.55+0.0336
90%CI:: 0.5164 < p < 0.5836
(i) Interpret your result in part (h).
Reject Ho because 0.55 is not in the 90% CI
(j) Do the conclusions in parts (g) and (i) agree with each other?
Yes
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Cheers,
Stan H.
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