Finding critical values: 1) 2-tailed test a=0.01 assume normal distribution applies. Critical z value? and How did you arrive at it?
The test we are considering is a two tailed test with
The total area of the two tails is 0.01.
The area of each tail is 0.01/2 = 0.005.
The graph is the probability density function of the standard normal distribution. The total area
under the curve is 1. In the diagram above, the area is divided in to three parts.
(1) The area from to -2.575829 is equal to 0.005
(2) The area from to 2.575829 is equal to 0.99
(3) The area from to is equal to 0.005
From this, it is clear that the critical values are and
If we round to 2 decimal places, the critical values are -2.58 and 2.58.
The area from to 2.575829 is equal to 0.995
How to find out the critical values?
Use the table of areas under the standard normal curve. From the table find the value of z such that the area from to z is 0.995. One can easily find that the value is 2.58(approximately).
2) Left-tailed test a=0.10 Critical z value?
Area of the left tail is 0.10
From the table, the area from to -1.28 is 0.10
So the critical value is -1.28
3) a=0.01; H 1 is p> 0.5...Critical z value?
This is a right tailed test.
Right tail area = Area from z to 0.01
Area from to z is 1-0.01=0.99
From tables z = 2.33
4) a=0.005; H1 is p not equal to (= w/slash thru) 45mm. Critical z value? Find value of test statistic z using z=p hat- p over square root of p times q divided by n.
Critical value
H1 is p not equal to 45
So it is a two tailed test
Area of two tails 0.005
Area of one tail = 0.005/2=0.0025
Area from to z is 1-0.0025=0.9975
Critical value is 2.81
The critical values are -2.81 and 2.81
5) Claim: less than half adults in US have carbon monoxide detectors. A survey of 1,005 adults resulted in 462 who have carbon monoxide detectors.
The level of significance is not given in the question. So let us take it as
The critical value of z is -1.645
So the decision rule is to reject the null hypothesis if z < -1.645.
The computed value of z is -2.555 and falls in the rejection region.
So we reject the null hypothesis and conclude that less than half of adults in US have carbon monoxide detectors.
6) Claim: more than 75% adults wear seat belt in front seat. Poll of 1,012 adults resulted in 870 who say theu wear seat belt in front seat. Find P-value using 0.05 significance level and state whether to reject null hypothesis or if failed to reject null hypothesis.
H0: p = 0.75
H1: p > 0.75
n= 1012
x= 870
Critical value of z = 1.645
Decision rule reject the null hypothesis if z > 1.645
The p-value is P(z > 8.058) =0.0000
The p-value is less than 0.05 and so we reject the null hypothesis.
Conclusion: z falls in the rejection region. Reject the null hypothesis.
7) Test statistic in a 2-tailed test is z=1.75.
If the level of significance is 0.05, then the critical value of z is 1.96.
So 1.75 does not fall in the rejection region. So fail to reject the null hypothesis.
8) With H1:p 0.777 test statistic is z= -2.95 Testing Claims about proportions...ID the null hypothesis alternative hypothesis test statistic P-value or critical value(s) conclusion about null and final conclusion about original claim. Use P-value method and the normal distribution s an approximation to the binomial distribution...
The question is not clear
For problems 9 & 11 I just need these mapped-out on how they were solved), but all just above apply to #10:
9) In a poll 745 randomly selected adults, 589 said it's morally wrong to not report all income on tax returns. Use a 0.01 significance level to test the claim that 75% of adults say it's morally wrong to not report all income on tax returns.
(I have the answer of... H0: p = 0.75 H1 : p = 0.75 Test statistic: z = -2.56. Critical values: z = +/- 2.575 P value: 0.0104 Fail to reject H0. Not enough evidence to reject claim that 75% of adults say it's morally wrong to not report all their income on tax return.)...BUT I don't know how it was arrived at. Please only map-out for me how it was solved.
Solution
n = 745
x = 589
z =
p-value = P
The p-value is not less than the critical value. So we fail to reject the null hypothesis.
The critical value of z is 2.575
Decision rule: Reject the null hypothesis if z < -2.575 or z > 2.575
Since z does not fall in the rejection region, we fail to reject the null hypothesis.
Conclusion:
75% of adults say it's morally wrong to not report all income on tax returns.
10) A test for marijuana usage. Among 300 tested subject results from 27 were wrong (either false positive or false negative). Use a 0.05 significance level to test the claim that less than 10% of the test results are wrong. Does this test appear to be good for most purposes?
H0: p=0.10
H1: p <
Critical value of z = -1.645
Decision rule: reject the null hypothesis if z < -1.645
n= 300
x=27
z does not fall in the rejection region
Fail to reject the null hypothesis.
11) When testing MI gas pumps for accuracy etc. a study found 1,299 of them were not pumping accurately (w/in 3.3 oz. when 5 gal. pumped) and 5,686 pumps were accurate. Use a 0.01 significance level to test claim that less than 20% of the gas pumps are inaccurate. From the position of a consumer does the rate appear to be low enough? (I have the answer of: H0: p = 0.20 H1: p< 0.20 Test statistic: z = -2.93 Critical value: z = -2.33 Pvalue: 0.0017 Reject H0 There is enough evidence for claim that less than 20% of MI gas pumps are inaccurate. % of inaccurate pumps should be low but this hypothesis test shows only that it appears to be less than 20%. The sample proportion of 0.186 suggests that too many pumps are inaccurate.) Testing Hypotheses...ID the null hypothesis alternative hypothesis test statistic P-value or critical value(s) conclusion about null hypothesis and final conclusion addressing the original claim. Use the P-value method.
Not Accurate = 1299
Accurate = 5686
Total = 1299 + 5686 = 6985
Proportion of not accurate pumps =
H0: p=0.20
H1:p<0.20
p-value =
The p-value is less than the significance level 0.01. So reject the null hypothesis.
There is evidence that less than 20% of the pumps are inaccurate.
12) Simple random sample of 50 adults and ea. person's red blood cell count (in cells per microliter) is measured. Sample mean is 5.23. The population standard deviation for red blood cell counts is 0.54. Use a 0.01 significance level to test the claim that sample is from a pop. with a mean less than 5.4. What do the results suggest about the sample group? Can you answer them for me within 36 hrs. from now? Rachel
n=50
H0:
Critical z = -2.326
Decision rule : Reject H0 if z < -2.326.
Conclusion: z does not fall in the rejection region. So we fail to reject the null hypothesis. The sample cannot considered as from a population with mean less than 5.4