1. In a random study of college students regarding marijuana use, 13 out o dents

Question

1. In a random study of college students regarding marijuana use, 13 out o dents admitted to occasionally smoking dents admitted to occasionally smoking marijuana. In this situation you want to use a marijuana while only 10 out of 35 females respon- confidence interval. If you did a significance test to see if the proportions for each group were different, the standard error used to compute the z statistie would be 2. Suppose you suspect that the average birthweight of golden retrievers is greater than the average birthweight of yellow labradors. You take an SRS of 24 newborn golden retrievers and find a the mean weight to be 1.2 pounds with standard deviation 0.35 pounds. You then take an SRS of 21 newborn yellow labradors and find the mean weight to be 1.1 pounds with standard deviation 0.4 pounds. If you do a siguificance test, the P-value will be between and

Explanation / Answer

Solution:-

1)

We want to use 95% confidence interval.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.3333

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.1135

z = (p1 - p2) / SE

z = 0.8514

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 0.85 or greater than 0.85.

Thus, the P-value = 0.3954

Interpret results. Since the P-value (0.3954) is greater than the significance level (0.05), we have to accept the null hypothesis.

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