Question

1.) A recent survey showed that high school girls average 110 text messages daily. The population standard deviation is 25 text messages. 1 In repeated samples of size n = 64 high school girls, the expected value of the sample mean is,

a 1.719 b 10.488 c 13.75 d 110

2.) In repeated samples of size n = 64 high school girls, the standard error of the sample mean is,

a 25 b 5 c 3.125 d 0.391

3.) In repeated samples of size n = 64 high school girls, the fraction of sample means that fall within ±5 messages from the population mean is,

a 0.8904 b 0.9260 c 0.9538 d 0.9824

4.) The margin of error for the middle interval which includes 95% of means from samples of size n = 64 is,

a 5.763 b 5.941 c 6.125 d 6.248

Answer 1

n = 64

According to central limit theorem, if sample size n is large (n>30) then sampling distribution of sample mean is approximately normally distributed with mean = and standard deviation =

1) Expected value of sample mean =

Expected value of sample mean =110

2) Standard error of sample mean

= 3.125

Standard error of sample mean is 3.125

3) Here we have to find               (add and subtract 5 from population mean 110)

                                       

                                       

                                       = P(z < 1.6) - P(z < -1.6)

                                       = 0.9452 - 0.0548                           (From statistical table of z values)

                                       = 0.8904

In repeated samples of size n = 64 high school girls, the fraction of sample means that fall within ±5 messages from the population mean is 0.8904

4) Margin of error (E):

Confidence level = c = 0.95

Where zc is z critical value for (1+c)/2 = (1+0.95)/2 = 0.975 is 1.96        (From statistical table of z values)

E = 1.96 * 3.125

   = 6.125

Margin of error = 6.125