3. A researcher wanted to determine if carpeted or uncarpeted rooms contain more

Question

3. A researcher wanted to determine if carpeted or uncarpeted rooms contain more bacteria. The table shows the results for the number of bacteria per cubic foot for both types of rooms. A normal probability plot and boxplot indicate that the data are approximately normally distributed with no outliers. Do carpeted rooms have more bacteria than uncarpeted rooms at the a=.01 level of significance? Carpeted 63 .5 12.1 13.6 14.5 10.5 6.5 12.3 Uncarpeted|2.9|6.852 18.6 9.9 1.1 |3.7|2.8 4. In randomized, double-blind clinical trials of a new vaccine, monkeys were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the first dose, 115 of 701 subjects in the experimental group experienced fever as a side effect. After the first dose, 68 of 609 of the subjects in the control group experienced fever as a side effect. 1) Construct a 95% confidence interval for the difference between the two population proportions, pl-p2.2) Construct a hypothesis test at 0.1 level of significance to see ifpl>p2

Explanation / Answer

Solution:-

6)

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

Null hypothesis: P < 0.30
Alternative hypothesis: P > 0.30

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.0324
z = (p - P) /

z = 2.31

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 2.31.

Thus, the P-value = 0.0104

Interpret results. Since the P-value (0.0104) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that he will have more than 30% of the votes in this coming election.

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