Question

In a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month. Use a 0.05 significance level to test the claim that 80%of patients stop smoking when given sustained care.
-Identify the Test statistic

-Identify the P-Value

for this Hypothesis Test

Answer 1

Answer:

We conclude that 80% of patients stop smoking when given sustained care.

Step-by-step explanation:

We are given that in a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month.

Let p = percentage of patients stop smoking when given sustained care.

So, Null Hypothesis, $H_0$ : p = 80%     {means that 80% of patients stop smoking when given sustained care}

Alternate Hypothesis, $H_A$ : p $\neq$ 80%     {means that different from 80% of patients stop smoking when given sustained care}

The test statistics that would be used here One-sample z test for proportions;

                        T.S. =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of patients who stop smoking when given sustained care = 80.2%

           n = sample of patients = 192

So, the test statistics  =  $\frac{0.802-0.80}{\sqrt{\frac{0.802(1-0.802)}{192} } }$

                                     =  0.07

The value of z test statistics is 0.07.

Also, P-value of the test statistics is given by the following formula;

                P-value = P(Z > 0.07) = 1 - P(Z $\leq$ 0.07)

                              = 1 - 0.52790 = 0.4721

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 80% of patients stop smoking when given sustained care.