Question

The local government is considering implementing a roads and public transport infrastructure upgrade project. Before they commit to this however, they would like to canvas public opinion to gauge community support for such a project. If it they are convinced that more than 60% of the community support the proposed upgrade project, 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. Download the data Survey results: yes yes yes no yes yes yes yes yes yes yes yes yes yes yes yes yes no yes yes no yes yes yes yes no no yes no yes yes no yes yes no yes yes yes yes yes no yes yes yes yes yes no yes yes yes yes yes yes yes yes no yes no no yes no yes no yes no no no no no no no yes yes yes no yes no yes yes no yes no yes yes no yes yes no yes yes no yes yes no no no yes yes yes yes yes yes yes yes no no yes yes yes yes The level of significance to be used in the test is α = 0.05. a)From the following options, select the correct null and alternate hypotheses for this test: A:H0: p < 0.6, Ha: p > 0.6 B:H0: p = 0.6, Ha: p ≠ 0.6 C:H0: p = 0.6, Ha: p > 0.6 D:H0: p = 0.6, Ha: p < 0.6 The correct null and alternate hypotheses for this test are: b)Calculate the test statistic (z) for this hypothesis test. Give your answer to 2 decimal places. z = c)Therefore, at a significance level of 0.05, the null hypothesis is . That is, you can state that there is to conclude that the of people who support the proposed upgrade project is 0.6. [1 point]

Answer 1

Here we have given that,

n=Total number of people selected in group= 110

x: number of people supported the proposed upgrade project =76 (i.e. count of 'yes' response from the provided data)

Now, we estimate the sample proportion as

=sample proportion of the community support the proposed upgrade project =

p=population proportion of the community support the proposed upgrade project= 0.60

The below mentioned necessary assumption is satisfied for this hypothesis test (one sample proportion test).

• The data are simple random sample values form the population.
• The response is binary i.e. either success/failure and the distribution is binomial.
• The mean n*p=110*0.60=66 and the variance = n*(1-p)= 110*0.60*(1-0.60)=26 are greater than 10, the binomial distribution can be approximated by the normal distribution.

(A)

Claim: To check whether the proportion of the community support the proposed upgrade project is more than 60% i.e. 0.60

The null and alternative hypothesis are as follows,

v/s

where p is the population proportion

This is the right one-tailed test.

Option C is correct.

(B)

Now, we can find the test statistic is as follows,

Z-statistics=

       =

         =1.95

The test statistics is 1.95.

Now we find the P-value,

= level of significance= 0.05

p-value=P(Z > z-statistics)

= 1- P( Z < 1.95)

             =1 - 0.9744   Using standard normal z table see the value corresponding to the z=1.95

=0.0256

The p-value is 0.0256

Decision:

Here p-value (0.0256) < 0.05

Conclusion:

We reject the Ho (Null Hypothesis)

There is sufficient evidence to support the claim the proportion of the community support the proposed upgrade project is more than 60% i.e. 0.60.