1 of 13 ID: MST.HT.TP.05.0010 [5 points] he local government is considering impl
ID: 2908119 • Letter: 1
Question
1 of 13 ID: MST.HT.TP.05.0010 [5 points] he local government is considering implementing a roads and public transport infrastructure upgrade project. Before they commit to this however, they would like to a vas public opinion to gauge com munity support for such a pro ect. If it they are convinced that more than 60% of the community support the proposed upgrade roject, then the government will commission the project The following sample was collected by asking a randomly selected group of 110 people whether or not they supported the proposed upgrade project. Survey resu Download the data yes no no yes yes no yes no no yes yes yes yes no yes no yes yes yes yes yes yes no no yes no yes yes yes no yes no no yes no yes yes yes yes yes yes| yes l no l yes | no | no | no | yes | yes | no : yes l yes | yes | yes | no | yes | no | yes | yes no yes yes no yes yes yes yesyes yes yes no yes yes no yes yes yes no yes no yes yes yes yes no yes yes yes no yes yes no yes yes yes yes yes yes yes no yes no no yes no yes yes yes yes The level of significance to be used in the test is o 0.05 a) From the following options, select the correct null and alternate hypotheses for this test: A: Ho: -0.6, H: p 0.6 B:Ho: D-0.6, H,: D>0.6 C: Ho: D-0.6, H: pExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.60
Alternative hypothesis: P > 0.60
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).
S.D = sqrt[ P * ( 1 - P ) / n ]
S.D = 0.04671
z = (p - P) / S.D
z = 3.64
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 3.64..
Thus, the P-value = less than 0.0001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.
Reject the null hypothesis,
That is you can stat that there is sufficient evidence to conclude that the opinion of people who support the proposed upgrade project is more than 60%.
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