Question

The table below contains the birth weights in grams of 26 African American babies born at BayState Medical Center in Springfield, Massachusetts in 1986. Compute a 95% confidence interval for birth weight.


Directions: Click on the Data button below to display the data. Copy the data into a statistical software package and click the Data button a second time to hide it.

Data

Weight
2508
2774
2927
2951
2910
2961
2960
3047
3030
3352
3416
3392
3477
3789
3857
1174
1666
1952
2146
2178
2307
2383
2406
2410
2476
2508

1. Find the point estimate for the birth weights. Round your answer to 2 decimal places.
   
   
   
2. Determine the value of tctc. Round your answer to 5 decimal places.
   
   
   
3. Find the margin of error for the confidence interval. Round your answer to 1 decimal place.
   
   
   
4. Construct the confidence interval for birth weights. Enter your answer as an open interval of the form (a,b) and round to the nearest integer.
   
   
   
5. Babies weighing less than 2500 grams are considered to be of low birth weight. Can you conclude that the average birth weight is greater than 2500 grams?
   • No, the entire confidence interval is below 25002500.
   • No conclusions can be drawn since the confidence interval contains 25002500.
   • Yes, the entire confidence is above 25002500.

Answer 1

Solution:

Given the data on the birth weights in grams of 26 African American babies born at BayState Medical Center in Springfield, Massachusetts in 1986 and we have to compute a 95% confidence interval for birth weight.

We have use R programming to test the hypothesis

> Weight = c(2508,2774,2927,2951,2910,2961,2960,3047,3030,3352,3416,3392,3477,

 + 3789,3857,1174,1666,1952,2146,2178,2307,2383,2406,2410,2476,2508)  1) Point estimate for the birth weights is given by the sample mean > mean(Weight) [1] 2729.115

So, the point estimate for the birth weights is 2729.11500

2) The value of t critical at 5% level of significance and 25 degree of fredom is

t ctc = 1.70814

3) Margin of error is given by

M.E = tctc*S.E

Where, S.E = standard error

> S.E = sd(Weight)/sqrt(26)

 > S.E [1] 124.8885 > M.E = 1.70814*S.E > M.E [1] 213.3

Hence, margin of error is 257.2

4) The the confidence interval for birth weights is given by

(mean - M.E, mean +M.E )

> mean(Weight)-M.E

[1] 2515.788 > mean(Weight)+M.E [1] 2942.442

So, confidence interval is

(2515.788, 2942.442)

Yes, we can conclude that the average birth weight is greater than 2500 grams, SInce the entire confidence is above 2500.