Question

The length of human pregnancies is approximately normally distributed with mean 266 days and standard deviation 16 days. Dr. Smith obtains a simple random sample of 10 of her patients and obtains the following results:
279 260 261 266 255 267 230 266 264 240
Construct a 90% confidence interval for the gestation period for all of Dr. Smith's patients. Interpret this interval. .

Answer 1

Answer:

90% confidence interval for the gestation period for all of Dr. Smith's patients = [250.48 , 267.12] .

Step-by-step explanation:

We are given that the length of human pregnancies is approximately normally distributed with Mean, $\mu$ = 266 days and  standard deviation, $\sigma$ = 16 days.

Let suppose, Z = $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ follows N(0,1)

    where, $Xbar$ = Sample mean = $\frac{279 +260 +261 +266+ 255+ 267+ 230+ 266+ 264+ 240}{10}$

                                                     = 258.8

                  n = sample size = 10

So, the 90% confidence interval for $\mu$ is given by ;

 P(-1.6449 < N(0,1) < 1.6449) = 0.90 {because at 10% level of significance z

                                                                table gives critical value of 1.6449}

P(-1.6449 < $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.6449) = 0.90

P(-1.6449*$\frac{\sigma}{\sqrt{n} }$ < $Xbar - \mu$ < 1.6449*$\frac{\sigma}{\sqrt{n} }$ ) = 0.90

P($Xbar - 1.6449*\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $Xbar + 1.6449*\frac{\sigma}{\sqrt{n} }$ ) = 0.90

So, 90% confidence interval for $\mu$ =  $[Xbar - 1.6449*\frac{\sigma}{\sqrt{n} } , Xbar + 1.6449*\frac{\sigma}{\sqrt{n} }]$

                                                        = $[258.8 - 1.6449*\frac{16}{\sqrt{10} } , 258.8 + 1.6449*\frac{16}{\sqrt{10} }]$

                                                        = [250.48 , 267.12]

Since, 266 lies inside this interval so we conclude that the mean gestation period for all of Dr. Smith's patients is 266 days.

And this interval states that we have 90% confidence in the above statement.