Respuesta :
Answer:
We conclude that less than thirty percent of the teen girls smoke to stay thin.
Step-by-step explanation:
We are given that the Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old).
After four years the girls were surveyed again. Sixty-three said they smoked to stay thin.
Let p = percentage of the teen girls who smoke to stay thin.
So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\geq[/tex] 30% {means that at least thirty percent of the teen girls smoke to stay thin}
Alternate Hypothesis, [tex]H_A[/tex] : p < 30% {means that less than thirty percent of the teen girls smoke to stay thin}
The test statistics that would be used here One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample % of teen girls who smoke to stay thin = [tex]\frac{63}{273}[/tex] = 0.231
n = sample of teen girls = 273
So, test statistics = [tex]\frac{0.231-0.30}{\sqrt{\frac{0.231(1-0.231)}{273} } }[/tex]
= -2.705
The value of z test statistics is -2.705.
Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test.
Since our test statistics is less than the critical value of z as -2.705 < -1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that less than thirty percent of the teen girls smoke to stay thin.
