Question

Hand sanitizer containing 70% ethanol (HL+) is recommended to prevent the transmission of the rhinovirus (RV) that causes the common cold. A study with 212 subjects was conducted. Subjects were assigned at random to either the HL+ group, which used hand sanitizer after hand washing, or a control group, which did not use hand sanitizer after hand washing. The number of subjects with and without RV infection in the two groups over the ten week study are below. RV Infection Yes No HL+ Group 49 67 Control Group 49 47 (a) (2 points) Is this an experiment or observational study? Why? (b) (2 points) What is the response variable? Is it qualitative or quantitative? (c) (5 points) Does the data suggest that the RV infection rate of subjects using hand sanitizer is significantly (a = 0.05) lower? Perform the appropriate procedure and include all the steps. (d) (2 points) Construct and interpret a 95% confidence interval to estimate the difference between the two proportions.

Answer 1

Answer:

a) Experiment

b) Proportion of subjects infected. Quantitative

c) Null hypothesis failed to be rejected. P-value=0.101.

There is not enough evidence to support the claim that the RV infection rate of subjects using hand sanitizer is significantly lower.

d) The  95% confidence interval for the difference between proportions is (-0.223, 0.047).

As we can see, the value 0 (meaning no difference between the proportions) is included in the interval. This can be interpreted the sam wat that the hypothesis test: at 95% confidence, we can not assurre that there is significant difference between the proportions.

Step-by-step explanation:

a) This is an experiment, as the groups are designed by the researcher. The individuals are then asigned randomly to one of both groups.

b) The response variable is the proportion of subjects infected. This variable is quantitative, as is a sum of positive cases divided by the total amount of subjects in the group.

c) This is a hypothesis test for the difference between proportions.

The claim is that the RV infection rate of subjects using hand sanitizer is significantly lower.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

The significance level is 0.05.

The sample 1 (HL+ group), of size n1=116 has a proportion of p1=0.4224.

$p_1=X_1/n_1=49/116=0.4224$

The sample 2 (Control group), of size n2=96 has a proportion of p2=0.5104.

$p_2=X_2/n_2=49/96=0.5104$

The difference between proportions is (p1-p2)=-0.088.

$p_d=p_1-p_2=0.4224-0.5104=-0.088$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{49+49}{116+96}=\dfrac{98}{212}=0.4623$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.4623*0.5377}{116}+\dfrac{0.4623*0.5377}{96}}\\\\\\s_{p1-p2}=\sqrt{0.0021+0.0026}=\sqrt{0.0047}=0.0688$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.088-0}{0.0688}=\dfrac{-0.088}{0.0688}=-1.28$

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=P(z$

As the P-value (0.101) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the RV infection rate of subjects using hand sanitizer is significantly lower.

d) We want to calculate the bounds of a 95% confidence interval.

For a 95% CI, the critical value for z is z=1.96.

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.4623*0.5377}{116}+\dfrac{0.4623*0.5377}{96}}\\\\\\s_{p1-p2}=\sqrt{0.0021+0.0026}=\sqrt{0.0047}=0.0688$

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.0688=0.1348$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = -0.088-0.1348=-0.223\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= -0.088+0.1348=0.047$

The  95% confidence interval for the difference between proportions is (-0.223, 0.047).

As we can see, the value 0 (meaning no difference between the proportions) is included in the interval. This can be interpreted the sam wat that the hypothesis test: at 95% confidence, we can not assurre that there is significant difference between the proportions.