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## You have been doing research for your statistics class on the prevalence of severe binge drinking among teens. You have decided to use 2011

Question

You have been doing research for your statistics class on the prevalence of severe binge drinking among teens. You have decided to use 2011 Monitoring the Future (MTF) data that have a scale (from 0 to 14) measuring the number of times teens drank 10 or more alcoholic beverages in a single sitting in the past 2 weeks.

a. According to 2011 MTF data, the average severe binge drinking score, for this sample of 914 teens, is 1.27, with a standard deviation of 0.80. Construct the 95% confidence interval for the true averse severe binge drinking score.

b. On of your classmates, who claims to be good at statistics, complains about your confidence interval calculation. She or he asserts that the severe binge drinking scores are not normally distributed, which in turn makes the confidence interval calculation meaningless. Assume that she or he is correct about the distribution of severe binge drinking scores. Does that imply that the calculation of a confidence interval is not appropriate? Why or why not?

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Mathematics
6 months
2021-07-25T03:13:25+00:00
2021-07-25T03:13:25+00:00 1 Answers
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## Answers ( )

Answer:

(1.218 ; 1.322)

the confidence interval is appropriate

Step-by-step explanation:

The confidence interval :

Mean ± margin of error

Sample mean = 1.27

Sample standard deviation, s = 0.80

Sample size, n = 914

Since we are using tbe sample standard deviation, we use the T table ;

Margin of Error = Tcritical * s/√n

Tcritical at 95% ; df = 914 – 1 = 913

Tcritical(0.05, 913) = 1.96

Margin of Error = 1.96 * 0.80/√914 = 0.05186

Mean ± margin of error

1.27 ± 0.05186

Lower boundary = 1.27 – 0.05186 = 1.218

Upper boundary = 1.27 + 0.05186 = 1.322

(1.218 ; 1.322)

According to the central limit theorem, sample means will approach a normal distribution as the sample size increases. Hence, the confidence interval is valid, the sample size of 914 gave a critical value at 0.05 which is only marginally different from that will obtained using a normal distribution table. Hence, the confidence interval is appropriate