und that he suspects can his 2.Mr Scientist has developed a compo affect the hei
ID: 3359006 • Letter: U
Question
und that he suspects can his 2.Mr Scientist has developed a compo affect the height of a growing individual. In order to test hypothesis, he injected the same dosage of the compound to each of the 40 growing guinea pigs in his treatmen" group and injected nothing to each of the other 90 growing guinea pigs in his "aontrol" group. Then, after six months, increase in body length in each of these treated/untreated 130 growing guinea pigs. The following is his results: he measured the Treatment group (sample size is 40): Sample mean increase in body length is 1.40 inches Sample standard deviation is o.20 inch Sample variance is o.04 inch control group (sample size is 90): Sample mean increase in body length is 1.30 inches Sample standard deviation is 0.30 inch Sample variance is 0.09 inch Use a significance level of 0.05, test the null hypothesis that the increases in body length of the two groups are the same against the alternative hypothesis that the increases in body length of the two groups are not the sane. Based on these information, the calculated z 1S (x1-x2) A. )+2.236 D. Not enough information given to answer this question E. None of the above B. -2.236 C. +50.000 [( x,:l.to 1.30 S.-.20 10 -1,9 1.96 s0Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 = 2
Alternative hypothesis: 1 2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample z-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.04472
z = [ (x1 - x2) - d ] / SE
z = 2.236
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a z statistic more extreme than -2.236; that is, less than -2.236 or greater than 2.236
Thus, the P-value = 0.025
Interpret results. Since the P-value (0.025) is less than the significance level (0.05), we cannot accept the null hypothesis.
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