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The mean height of a random sample of 47 students who take part in athletic activities...

The mean height of a random sample of 47 students who take part in athletic activities at NRDC is 175cm with a standard deviation of 5cm while a random sample of 132 students who showed no interest in athletics had a mean of 170cm and a standard deviation of 7cm. (a) Construct a 99% confidence interval for the difference in the mean heights of the two groups of students. (b) Are students who take part in athletic activities taller? Test at the 1% level of significance

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Answer #1

(a)

Confidence interval for the difference in means is given as:

Upper limit = X1-X2 + Marqin of error

Lower limit = X1-X2-Margin of error

Margin of error- ta/2.df SE

SE is given by by :

ni+n2-2

(47-1)52+(132-1)7211 V 47+132-2 47132

SE = 0.9504

ta 2.df-to toob ії3.242-2.62 01/2.113.242

Margin of error - ta/2.df SE 2.62 * 0.9504 2.49

99% confidence interval for the difference in the mean heights of the two groups of students :

Upper limit- 175-170+2.49-52.49 7.49

Lower limit-175-170-2.49-5-2.49 2.51

b)

We conduct T-test for two Means :

The provided sample means are shown below: X 175 X2170 Also, the provided sample standard deviations are 81-5 82-7 and the sample sizes are n1 = 47 and n2 132. (1) Null i/ld Aliornain代冫//ypothosss The following null and alternative hypotheses need to be tested: This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.Based on the information provided, the significance level is a 0.01, and the degrees of freedom are df - 113.242. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal Hence, it is found that the critical value for this right-tailed test is tc-2.36, for α-0.01 and df- 113.242. The rejection region for this right-tailed test is R- ft:t>2.36 Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: n1+n2-2 2 175 17O = 5.261 (47 45-132 321) 1 1 (T.) 47+132-2 47 132 (4) Decision about the null hypothesis Since it is observed that t-5.261 > tc-2.36, it is then concluded that the null hypothesis is rejectedUsing the P-value approach: The p-value is p 0, and since p 0< 0.01, it is concluded that the null hypothesis is rejected (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim ulation mean μ1 is greater than μ 2, at t 0.01 significance level.Hence, we conclude that Students who take part in athletic activities are actually taller at 1% level of significance.

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