Question 1042154: 600 people randomly polled from a city showed 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?
2. Estimating the mean height of men in your city.
(a) Select a random sample of 25 men from your city. Record the 25 their heights in inches.
Estimating the mean height of men in your city.
(a) Select a random sample of 25 men from your city. Record the 25 their heights in inches.
65, 65, 64, 65, 72, 71, 73, 60, 60, 71, 66.5, 62.5, 54, 70,
63, 64, 71, 73, 60, 62, 63.5, 57, 60, 64, 65
(b) Describe how the sample is selected.
(c) Make a stemplot for the data.
(d) Find the mean and standard deviation of the 25 men's heights
= ________ inches
s = _________ inches
(e) df =
(f) Give a 90% confidence interval for the mean height of all men in your city.
(g) Interpret your result in part (f).
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The sample proportion is 0.55
Ho: There is no majority who support gun control, or p< = 0.50
Ha: There is a majority, p>0.50.
alpha=0.05, and the sample was randomly chosen among the population.
Test statistic is a z-test for 1 sample proportion, with critical value of z>1.645
z=(p obs-p hypoth)/sqrt((p)(1-p)/n))
z=(0.55-0.50)/sqrt(0.55)*0.45)/600)
=0.05/0.0203
z=2.46
p-value is 0.007
This is in the rejection region.
The null hypothesis is rejected and the conclusion is that a majority support gun control.
90% CI has an interval of z(0.90)*Std error.
That is 1.645*0.0203=0.0333
The interval is around 0.55
(0.517,0.583)
every value in that interval could be consistent, and 0.50 is not in the interval, so the above is confirmed.
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This is a simple random sample.
5|47
6|5545006.52.534023.5045
7|213101
Mean is 64.86 inches
sd=5.088 inches
df=24
t 0.90,df=24=1.711,
standard error is s/sqrt(n)=5.088/5=1.018
90% CI is mean +/- t from above* SE
t*SE=1.711*1.018=1.74
The interval is 64.86+/-1.74, or (63.12,66.60)
That means that we are 90% confident that the true mean, which we do not know, lies in the above interval.
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