Question

In a sample of 51 babies, the mean weight is 21 pounds and the standard deviation is 4 pounds.

1. Calculate a 95% confidence interval for the true mean weight of babies. Suppose we are interested in testing if the true mean weight of babies is 19.4 vs the alternative that it is not 19.4 with an alpha level of .05. Would this test be significant? Explain your answer.

Perform the t test and use a t-table to get the p-value

Answer 1

a)

t critical value at 0.05 level with 50 df = 2.009

95% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

21 - 2.009 * 4 / sqrt(51) < < 21 + 2.009 * 4 / sqrt(51)

19.87 < < 22.13

95% CI is ( 19.87 , 22.13 )

b)

H0: = 19.4

Ha: 19.4

Test statistics

t = - / S / sqrt(n)

= 21 - 19.4 / 4 / sqrt(51)

= 2.86

This is test statistics value.

From T table, with t = 2.86 and df = 50

p-value = 0.0062

Since p-value < 0.05 level, we reject the null hypothesis.

We conclude at 0.05 level that we have sufficient evidence to support the claim.