Question

Suppose an investigator takes a random sample of n = 50 birth weights from several teaching hospitals located in an inner-city neighborhood. In her random sample, the sample mean x is 3,150 grams and the standard deviation is 250 grams.

1. (a) Calculate a 95% confidence interval for the population mean birth weight in these hospitals.

2. (b) ThetypicalweightofababyatbirthfortheUSpopulationis3,250grams.Theinvestigatorsus- pects that the birth weights of babies in these teaching hospitals is different than 3,250 grams, but she is not sure if it is smaller (from malnutrition) or larger (because of obesity prevalence in

   mothers giving birth at these hospitals). Carry out the hypothesis test that she would conduct.

Please could you give details. I have done the question but I want to make sure I got it right.

Thank you !

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 3150

Population standard deviation =    = 250

Sample size = n = 50

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 250 /  50 )

= 69.30

At 95% confidence interval estimate of the population mean is,

- E < < + E

3150 - 69.30 <   < 3150 + 69.30

3080.70 <   < 3219.30

( 3080.70 , 3219.30 )

The 95% confidence interval of the population mean is : ( 3080.70 , 3219.30 )

(b)

Given that ,

= 3250

= 3150

= 250

n = 50

The null and alternative hypothesis is ,

H₀ :   = 3250

H_a :    3250

This is the two tailed test .

Test statistic = z

= ( - ) / / n

= ( 3150 - 3250) / 250 / 50

= -2.828

The test statistic = -2.828

P - value = 2 * P (Z < -2.828 )

= 2 * 0.0023

= 0.0046

P-value = 0.0046

= 0.05

0.0046 < 0.05

P-value <

Reject the null hypothesis .

There is sufficient evidence to test the claim.