How can complex fractions be used to solve fractions that include ratios
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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.
<u><em>Let p = percentage of the teen girls who smoke to stay thin.</em></u>
So, Null Hypothesis, : p
30% {means that at least thirty percent of the teen girls smoke to stay thin}
Alternate Hypothesis, : p < 30% {means that less than thirty percent of the teen girls smoke to stay thin}
The test statistics that would be used here <u>One-sample z proportion statistics</u>;
T.S. = ~ N(0,1)
where, = sample % of teen girls who smoke to stay thin =
= 0.231
n = sample of teen girls = 273
So, <u><em>test statistics</em></u> =
= -2.705
The value of z test statistics is -2.705.
<em>Since, in the question we are not given the level of significance so we assume it to be 5%. </em><em>Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test.</em><em> </em>
<em>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 </em><em><u>we reject our null hypothesis</u></em><em>.</em>
Therefore, we conclude that less than thirty percent of the teen girls smoke to stay thin.