Question

The number of miles traveled by birds is known to be normally distributed. A random sample of 43 birds results in a mean distance of 2.64 miles with a standard deviation of 0.15 miles. You've been asked to construct a 95% confidence interval for the mean number of miles traveled. Based on this you would use the table. none of the above. Ttable Ftable Z table Question 16 1 pts Consider the following competing hypotheses: HON= 0, HAU# 0. The value of the test statistic is z=-1.46. If we choose a 1% significance level, what is the correct decision and conclusion? Do not reject the null hypothesis; we can conclude that the population mean is significantly different from zero. Reject the null hypothesis; we cannot conclude that the population mean is significantly different from zero. Do not reject the null hypothesis; we cannot conclude that the population mean is significantly different from zero. Reject the null hypothesis; we can conclude that the population mean is significantly different from zero.

Answer 1

Question N0 15

Given

Mean (Xabr)=2.64 Miles

Standard Deviation ($\sigma$)=0.15 miles

Sample size = 43

For 95% confidence level Z=1.96 (Refer standard normal table)

We have

Z = (X - Xbar)/$\sigma$

1.96 = (X- 2.64)/0.15

X = 1.96*0.15 + 2.64 =2.93

Therefore for 95% confidence level mean =2.93

Correct Answer is

Z table

Note :

(a) T -table shows probablity under probablity density funtion.

(b) F-table shows a probability density function that is used especially in analysis of variance.