Question

Harper’s Index claims that 23% of American are in favor of outlawing cigarettes. You decide to test this claim and ask a random sample of 200 Americans whether they are in favor of outlawing cigarettes. Out of 200 Americans, 54 are in favor. Use critical-value or p-value approach and a significance level of 0.05. Conduct a hypothesis test to determine whether there is enough evidence to reject the claim.

Answer 1

Answer:

There is enough evidence not to reject the claim.

Step-by-step explanation:

H0: The population proportion of Americans in favor of cigarette outlawing is 0.23.

Ha: The population proportion of Americans in favor of cigarette outlawing is not 0.23.

Critical value approach

z = (p' - p) ÷ sqrt[p(1 - p) ÷ n]

p' is sample proportion = 54/200 = 0.27

p is population proportion = 23% = 0.23

n is number of Americans sampled = 200

z = (0.27 - 0.23) ÷ sqrt[0.23(1 - 0.23) ÷ 300] = 0.04 ÷ sqrt[8.855×10^-4] = 0.04 ÷ 0.0298 = 1.34

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.

Critical value at 0.05 significance level is 1.96

For a two-tailed test, the region of no rejection of H0 lies between -1.96 and 1.96

Conclusion:

Fail to reject H0 because the test statistic 1.34 falls within the region bounded by -1.96 and 1.96.

There is enough evidence not to reject the claim because the claim is contained in the null hypothesis (H0).

Answer 2

Answer:

No, there is not enough evidence to reject the claim.

Step-by-step explanation:

We are given that Harper’s Index claims that 23% of American are in favor of outlawing cigarettes. To test this claim a random sample of 200 Americans are asked whether they are in favor of outlawing cigarettes. Out of 200 Americans, 54 are in favor.

Let Null Hypothesis, $H_0$ : p = 0.23 {means that % of Americans who are in favor of outlawing cigarettes is same as Harper’s Index claim of 23%}

Alternate Hypothesis, $H_1$ : p $\neq$ 0.23 {means that % of Americans who are in favor of outlawing cigarettes is different from Harper’s Index claim of 23%}

The z-test statistics we will use here is One sample proportion test ;

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

where, p = % of Americans who are in favor of outlawing cigarettes based  on the Harper’s Index claim = 23%

       $\hat p$ = % of Americans who are in favor of outlawing cigarettes based on a sample of 200 Americans = $\frac{54}{200}$ = 27%

      n = sample of Americans = 200

So, test statistics = $\frac{0.27 - 0.23}{\sqrt{\frac{0.27(1-0.27)}{200} } }$

                            = 1.274

Therefore, the z test statistic is 1.274 .

Now, at 0.05 significance level the critical value of z in z table is given as 1.96. Since our test statistics is less than the critical value of t which means our test statistics will not lie in the rejection region. So, we have insufficient evidence to reject our null hypothesis.

Therefore, we do not have enough evidence to reject the claim and conclude that % of Americans who are in favor of outlawing cigarettes is same as Harper’s Index claim of 23%.