A random sample of eight students participated in a psychological test of depth

Question

A random sample of eight students participated in a psychological test of depth perception. Two markers, one labeled A and the other B, were arranged at a fixed distance apart at the far end of the laboratory. One by one the students were asked two judge the distance between the two markers at the other end of the room. The sample data (in feet) were as follows: 1. 2.1, 2.2, 2.6, 2.3, 1.8, 2.3,2.4, 2.5 At alpha 0.0s level of significance, test if the mean distance is more than 2 feet. and Confidence Interval Vn rt State the null and alternative hypothesis. 4 points State your conclusion for your hypothesis test Give the 95% Confidence Interval. Give a conclusion for the confidence interval test. 4 points 4 points 4 points

Explanation / Answer

(a) Null HYpothesis : H0 : Mean distance is equal to two feet. = 2 feet.

Alternative Hypothesis : Ha : Mean distance is more than two feet. > 2 feet

(b) We will perform t - test here. But, first we will find

mean of the sample xbar = 2.2788

standard deviation of the sample s = 0.250

so standard error of the sample se = s/sqrt(n) = 0.25/ sqrt(8) = 0.0884

Test Statistic

t = (xbar - 2)/ se = (2.2788 - 2)/ 0.0884 = 3.154

critical value of t for dF = 7 and alpha = 0.05

t7,0.05 = 1.895

so here t > tcritical so we reject the null hypothesis and can say that mean distance is more than 2 feet.

95% confidence interval is +- E

where E = t* * s/sqrt(n) = 1.895 * 0.0884 = 0.1675

so 95% confidence interval= 2.2788 +- 0.1675 = (2.1113, 2.4463)

(d) as the confidence interval doesn't posses the value of 2 feet so we can reject the null hypothesis and can conclude that mean average distance is 2 feet.

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