Answer to Question #107894 in Statistics and Probability for phillip

Question

In a random sample of 300 industrial accidents, it was found that 183 were due at least in part to
unsafe working conditions. Use the level of significance  D 0:05 to decide whether this supports
the claim that 65 percent of such accidents are due at least in part to unsafe working conditions.
Which of the following statements is/are correct ?
A: H0
: p D 0:65; H1
: p > 0:65
B: The value of the test statistic is z D 1:45.
C: Critical value D 1:645 and the p-value D 0:0735.
D: H0
is rejected at 5% but not at 1% level of significance.
(1) B and C
(2) Only B
(3) A and D
(4) Only D

Answer 1

In a random sample of $N=300$ industrial accidents, it was found that $X=183$ were due at least in part to unsafe working conditions. Use the level of significance of $\alpha=0.05$ to decide whether this supports the claim that 65% of such accidents are due at least in part to unsafe working conditions.

The following information is provided: The sample size is $N=300,$ the number of favorable cases is $X=183,$ and the sample proportion is $\bar{p}=\dfrac{X}{N}=\dfrac{183}{300}=0.61,$ and the significance level is $\alpha=0.05.$

The following null and alternative hypotheses need to be tested:

$H_0:p=0.65$

$H_1:p>0.65$

This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=1.645.$

The rejection region for this right-tailed test is $R=\{z:z>1.645\}$

The z-statistic is computed as follows:

$$Z={\bar{p}-p_0\over \sqrt{p_0(1-p_0)/N}}={0.61-0.65\over \sqrt{0.65(1-0.65)/300}}\approx-1.452546$$

Since it is observed that $z=-1.452546<1.645=z_c,$ it is then concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than $p_0,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=1-0.073112=0.926888,$ and since $p=0.926888>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis $H_0$ is not rejected. Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than $p_0,$ at the $\alpha=0.05$ significance level.

There is no correct statement among A-D.