1)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.7
Alternative Hypothesis, Ha: p ≠ 0.7
Rejection Region
This is two tailed test, for α = 0.05
Critical value of z are -1.96 and 1.96.
Hence reject H0 if z < -1.96 or z > 1.96
Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.6403 - 0.7)/sqrt(0.7*(1-0.7)/1165)
z = -4.45
P-value Approach
P-value = 0
As P-value < 0.05, reject the null hypothesis.
There is sufficient evidence to conclude the claim
2)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.25
Alternative Hypothesis, Ha: p < 0.25
Rejection Region
This is left tailed test, for α = 0.01
Critical value of z is -2.33.
Hence reject H0 if z < -2.33
Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.245 - 0.25)/sqrt(0.25*(1-0.25)/200)
z = -0.16
P-value Approach
P-value = 0.4364
As P-value >= 0.01, fail to reject null hypothesis.
There is not sufficient evidence to conclude the claim
please show work, thank you in advance! 14 A government watchdog association claims that 70% of...
A government watchdog association claims that 70% of people in the U.S. agree that the government is inefficient and wasteful. A government agency wants to test this claim. In a random sample of 1165 people in the U.S., 746 agreed with that view. At a 95% confidence level, is there enough evidence to reject the association's claim? (Please show all work) a.) b.) c.) d.) e.)
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please help with both, thank you very much! Assume that the weights of adult women are normally distributed, and that the population standard deviation of weights of all women is o = 29 lb. A researcher claims that the weights of supermodels vary less than the weights of women in general. To test the claim, a random sample of nine supermodels is selected, with data (in pounds) shown as below: 125 119 128 128 119 127 105 123 115 Using...
PLEASE HELP ME OUT WITH BOTH QUESTIONS!!! it is greatly appreciated!! thank you so much. no need to show the work. You wish to test the claim that the average IQ score is less than 100 at the .005 significance level. You determine the hypotheses are: Ho:u= 100 H1:4 < 100 You take a simple random sample of 72 individuals and find the mean IQ score is 96.7, with a standard deviation of 15.5. Let's consider testing this hypothesis two...