Question

Part 5 of 6 In a simple random sample of 95 families, 70 had one or more pets at home. Can you conclude that the proportion of families with one or more pets differs from 0.67 Determine the parameter to be tested. The parameter to be tested is the population proportion Part: 5/6 Part 6 of 6 Determine the test to be performed. We have a simple random sample. The population (Choose one) at least 20 times as large as the samp The items in the population (Choose one) divided into two categories. The values nP, and (-P) (Choose one) V both at least 10. The correct decision is to (Choose one) the 2-test for the population proportion

Answer 1

Here we have given that,

n=number of families = 95

x= number of families had one or more pets at home=70

Now, we estimate the sample proportion $\small \hat p$ as

$\small \hat p$ =sample proportion of families had one or more pets at home = $\small \frac {x}{n}=\frac {70}{95}=0.737$

Claim: To check whether the population proportion of families with one or more pets differs from 0.60.

Here, The parameter to be tested is as follows,

The parameter to be tested is the population proportion of families with one or more pets p =0.60

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

1. we have a simple random sample.
2. The population is at least 20 times as large as the sample.
3. The items in the population are divided into two categories.
4. the values np=95*0.60= 57 and n(1-p)=95*(1-0.60)=38 are both at least 10
5. The correct decision to use the z-test for the population proportion.

The null and alternative hypothesis are as follows,

$\small Ho: p =0.60$

Versus

$\small H1: p \neq 0.60$

where p is the population proportion of families with one or more pets =0.60

This is the two-tailed test.

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

Z-statistics=$\small \frac{\hat p - p} {\sqrt \frac{p*(1-p)}{n}}$

= $\small \frac{0.737 - 0.6} {\sqrt \frac{0.6*(1-0.6)}{95}}$

=2.73

The test statistics is 2.73

Now we find the P-value,

p-value=2*( P(Z > z-statistics))

               =2* ( 1- P( Z < 2.73))

                =2* ( 1 - 0.99683) Using standard normal z table see the value corresponding to the z=2.73

                =2*0.00317

=0.0063

The p-value is 0.0063

Decision:

$\small \alpha$ = level of significance= 0.05

Here p-value (0.0063) less than (<) 0.05 $\small \alpha$

Conclusion:

We reject the Ho (Null Hypothesis)

There is sufficient evidence to support the claim the population proportion of families with one or more pets differs from 0.60.