A random selection of volunteers at a research institute have been exposed to a

Question

A random selection of volunteers at a research institute have been exposed to a typical cold virus. After they started to have cold symptoms, 25 of them were given multivitamin tablets daily which contain 3 grams of vitamin C and various other vitamins and minerals. The remaining 25 volunteers were given placebo tablets. For each individual, the length of time taken to recover from cold is recorded. At the end of the experiment following data are obtained: Days to recover from cold Treated with7.1, 5.9, 8.1, 2.3, 8.2, 5.8, 2.8, 5.6, 4.4, 6.8, 4.6, 6.5, 6.2, 2.9, 3.6, 2.9, 5.1, 5.4, 5.5, multivitamin 4.7, 7, 5.3. 1.5, 3, 1.5 Treated wth 4.5, 6.8, 1.7, 4.5, 4.7, 4.6, 2, 3.3, 7.7, 2.2, 4.3, 5.2, 7, 2.9, 2.7, 1.2, 5, 4.1, 5.2, 3.4, placebo 6.4, 5.4, 6, 5.4, 1.4 It is known that the population standard deviation of recovery time from cold is 1.8 days when treated with multivitamin, and the population standard deviation of recovery time from cold is 1.5 days when treated with placebo tablets. It is also known that both populations are approximately normally distributed. The researchers claim that the mean recovery time, of the patients treated with 1' multivitamin is not equal to the mean recovery time 2, of the patients who are treated with placebo tablets. At the 0.05 level of significance, is there enough evidence to support this claim? Perform a two- tailed test. Then fill in the table below Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) The null hypothesis:

Explanation / Answer

The statistical software output for this problem is:

Two sample Z hypothesis test:
1 : Mean of Treated with Multivitamin (Std. dev. = 1.8)
2 : Mean of Treated with placebo (Std. dev. = 1.5)
1 - 2 : Difference between two means
H0 : 1 - 2 = 0
HA : 1 - 2 0

Hypothesis test results:

Hence,

The null hypothesis: 1 = 2

The alternative hypothesis: 1 2

Type of test statistic: z

test statistic = 1.289

P - value = 0.197

No

Difference n1 n2 Sample mean Std. err. Z-stat P-value 1 - 2 25 25 0.604 0.46861498 1.2889046 0.1974

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