Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc or placebo lozenges every 2 to 3 hours until their cold symptoms were gone (Prasad et al., 2000). Twenty-five participants took zinc lozenges, and 23 participants took placebo lozenges. The mean overall duration of symptoms for the zinc lozenge group was 4.5 days, and the standard deviation of overall duration of symptoms was 1.6 days. For the placebo group, the mean overall duration of symptoms was 8.1 days, and the standard deviation was 1.8 days. a. Calculate a 95% confidence interval for the mean overall duration of symptoms if everyone in the population were to take the zinc lozenges.

b. Calculate a 95% confidence interval for the mean overall duration of symptoms if everyone in the population were to take the placebo lozenges.

c. On the basis of the intervals computed in parts (a) and (b) and/or the picture you constructed in part (c), is it reasonable to conclude generally that taking zinc loz-enges reduces the overall duration of cold symptoms more than if taking a placebo? Explain why you think this is or is not an appropriate conclusion.

d. In their paper, the researchers say that they checked whether it was reasonable to assume that the data were sampled from a normal curve population and decided that it was. How is this relevant to the calculations done in parts (a) and (b)?

a. (3.83952, 5.16049)

b. (7.32, 8.88)

c. The difference in the range of the confidence intervals suggest that taking the zinc loz-enges reduces the overall duration of cold symptoms

d. The assumptions required for the validity validity of a t-test include that the data is sampled from a source that have data that are normally distributed

Step-by-step explanation:

The given parameters are;

The number of participants that took zinc lozenges, n₁ = 25

The number of participants that took a placebo, n₂ = 23

The mean duration of the symptoms for the zinc lozenge group, = 4.5 days

The standard deviation of overall duration, s₁ = 1.6 days

The mean duration of the symptoms for the placebo group, = 8.1 days

The standard deviation, s₂ = 1.8 days

a. The 95% confidence interval if everyone in the population were to take zinc lozenges is given as follows;

n₁ = 25, the degrees of freedom (df) = n₁ - 1 = 24

The t-value for 95% confidence interval, with df = 24, with t = 2.064

Therefore, we have;

= 4.5 ± 0.66048

= (3.83952, 5.16049)

b. The 95% confidence interval if everyone in the population were to take the placebo is given as follows;

n₂ = 23, the degrees of freedom (df) = n₂ - 1 = 22

The t-value for 95% confidence interval, with df = 22, with t = 2.074

Therefore, we have;

= 8.1 ± 0.7784

= (7.32, 8.88)

c. Based on the difference in the range of the 95% confidence interval for the mean duration of cold symptoms of the group that take zinc lozenges, (3.83952, 5.16049), and the mean of the group that are on a placebo, (7.32, 8.88), there is sufficient statistical evidence to suggest that there is a difference between the mean and taking zinc lozenges reduces the overall duration of cold symptoms more than if taking placebo

d. The assumptions required for validity when carrying out a t-test include;

i) The measurement scale is ordinal or continuous

ii) The sample is a simple random sample

iii) The data gives a normal distribution curve

iv) The sample size is reasonably large

v) The variance are homogeneous

Among these assumptions, the data must be randomly sampled from the population of interest and that the data variables follow a normal distribution

The common assumptions made when doing a t-test include those regarding the scale of measurement, random sampling, normality of data distribution, adequacy of sample size and equality of variance in standard deviation

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

step-by-step explanation:

no solution.