i am using minitab to solve the problem.
steps :-
stat
basic statistics
2
proportion
select summarized data from the drop down menu
in
sample 1 type 40 in number of events , in number of trials type
120
in
sample 2 type 55 in number of events , in number of trials type
124
tick perform hypothesized test
options
in
confidence level type 90
in
hypothesized difference type 0
select the alternative hypothesis as difference < hypothesized
difference
in
method select use the pooled estimate of the proportion
ok
ok.
*** SOLUTION ***
a). z = -1.77
b). p-value = 0.039
c). Reject the null hypothesis. There is enough evidence to support the claim that the proportion of population 1 with the characteristics of interest is less than the proportion of population 1 with the characteristic of interest.
[ p value = 0.039 < 0.10 (alpha) ]
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- The sample information ort Sample 1 Sample 2 less than the proport ny = 120...
the accompanying sample information conduct the appropriate hypothesis test using the p-value i The sample information -X Sample 1 Sample 2 proport less than the proportion of pe n = 120 X1 = 40 n2 = 124 x2 = 55 Print Done Given the null and alternative hypotheses below, a level of significance a=0.1, together with the accompanying sample in hypothesis Ho: P2P2 HAPA <P2 Click on the icon to view the sample information Determine the value of the test...
Given the null and alternative hypotheses below, a level of significance a = 0.1, together with the accompanying sample information conduct the appropriate hypothesis test using the p-value approach. What conclusion would be reached concerning the null hypothesis? Ho: P1 = P2 HA:21 #P2 Х The sample information Click on the icon to view the sample information. Determine the value of the test statistic. Sample 1 Sample 2 Z= (Round to three decimal places as needed.) ny = 178 X1...
Use the following information to complete steps (a) through (d) below. A random sample of ny = 135 individuals results in xy = 40 successes. An independent sample of n2 = 150 individuals results in x2 = 60 successes. Does this represent sufficient evidence to conclude that P, <P2 at the a = 0.10 level of significance? (a) What type of test should be used? A. A hypothesis test regarding the difference between two population proportions from independent samples. B....
> Use the following information to complete steps (a) through (d) below. A random sample of n = 110 individuals results in xy = 40 successes. An independent sample of n2 = 160 individuals results in X2 = 60 successes. Does this represeot sufficient evidence to conclude that p, <P2 at the a= 0.01 level of significance? (a) What type of test should be used? O A. A hypothesis best regarding the difference between two population proportions from dependent samples....
The options for the final part
are A.(less than/greater than) B.(Fail to reject, Reject)
C.(sufficent/insufficent)
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3. Based on a random sample of 500 measurements, the sample proport Consider the test of hypothesis: Но: π-0.30 На: п < 0.30 with a-0.05 Calculate the test statistic z, the P-value, and make a decision about Ho. a. b. Construct a 95% confidence interval for the true population proportion Assuming the same confidence level as in the population proportion question (part b above), what sample size would be required to reduce the margin of error of the confidence interval...
14. Use the following information to complete steps (a) through (d) below. A random sample of n = 135 individuals results in x1 = 40 successes. An independent sample of n2 = 140 individuals results in X2 = 60 successes. Does this represent sufficient evidence to conclude that p1 <P2 at the a=0.05 level of significance? (a) What type of test should be used? O A. A hypothesis test regarding the difference between two population proportions from independent samples. OB....
Options for "The P-value is" Greater than. Less
than....... so reject or fail to
reject...... There is sufficient or nonsufficient
evidence
he Because the confidence interval limit
include
do not include
does
does not
appear to be a significant difference between the two
proportions. There
is not sufficient
is sufficient
evidence to warrant rejection of the claim that men and
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benefits as a job characteristic. The sample mean rating was 4.06,
and the sample standard deviation was 0.6. Test at the 1%
significance level the null hypothesis that the population mean
rating is at most 4 against the alternative that it is larger than
4.
What are the null and alternative hypotheses for this test?
For this...
You wish to test the following claim (H1H1) at a significance level of α=0.02α=0.02. Ho:p1=p2Ho:p1=p2 H1:p1<p2H1:p1<p2 You obtain 33.7% successes in a sample of size n1=718n1=718 from the first population. You obtain 42.5% successes in a sample of size n2=414n2=414 from the second population. 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...