In: Math
1. Out of
500 people sampled, 140 received flu vaccinations this year. Based
on this, construct a 95% confidence interval for the true
population proportion of people who received flu vaccinations this
year.
Give your answers as decimals, to three places
< p <
2. You are
conducting a study to see if the proportion of men over 50 who
regularly have their prostate examined is significantly more than
0.68. You use a significance level of α=0.05α=0.05.
H0:p=0.68H0:p=0.68
H1:p>0.68H1:p>0.68
You obtain a sample of size n=432n=432 in which there are 314
successes.
What is the test statistic for this sample? (Report answer accurate
to three decimal places.)
test statistic =
What is the p-value for this sample? (Report answer
accurate to four decimal places.)
p-value =
The p-value is... less than (or equal to) ααgreater than αα
This test statistic leads to a decision to... reject the nullaccept
the nullfail to reject the
null
As such, the final conclusion is that...
There is sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is more than 0.68.
There is not sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is more than 0.68.
The sample data support the claim that the proportion of men over 50 who regularly have their prostate examined is more than 0.68.
There is not sufficient sample evidence to support the claim that the proportion of men over 50 who regularly have their prostate examined is more than 0.68.
The effectiveness of a blood-pressure drug is being
investigated. An experimenter finds that, on average, the reduction
in systolic blood pressure is 47.3 for a sample of size 1066 and
standard deviation 6.8. Estimate how much the drug will lower a
typical patient's systolic blood pressure (using a 95% confidence
level).
Enter your answer as a tri-linear inequality accurate to one
decimal place (because the sample statistics are reported accurate
to one decimal place).
1. The statistical software output for this problem is:
One sample proportion summary confidence
interval:
p : Proportion of successes
Method: Standard-Wald
95% confidence interval results:
Proportion | Count | Total | Sample Prop. | Std. Err. | L. Limit | U. Limit |
---|---|---|---|---|---|---|
p | 140 | 500 | 0.28 | 0.020079841 | 0.24064424 | 0.31935576 |
Hence,
95% confidence interval:
0.241 < p < 0.319
2. The statistical software output for this problem is:
One sample proportion summary hypothesis
test:
p : Proportion of successes
H0 : p = 0.68
HA : p > 0.68
Hypothesis test results:
Proportion | Count | Total | Sample Prop. | Std. Err. | Z-Stat | P-value |
---|---|---|---|---|---|---|
p | 314 | 432 | 0.72685185 | 0.022443344 | 2.0875611 | 0.0184 |
Hence,
Test statistic = 2.088
p - Value = 0.0184
The p-value is less than (or equal to) α
Reject the null hypothesis
The sample data support the claim that the proportion of men over 50 who regularly have their prostate examined is more than 0.68.