Question

5. A researcher examined the folklore that women can predict the sex of their unborn child better than chance (50-50) would suggest. She asked 104 pregnant women to predict the sex of their unborn child, and 57 guessed correctly. Using this data, the researcher conducted a test at 5% significance level.

a) What is the parameter of interest?
b) State a null ?0 and alternative hypothesis ??.
c) What is the observed value?
d) Confirm conditions for the observed value to be distributed as normal.
e) Calculate the z-statistic.
f) For what values of ?̂ can the researcher reject the null?
g) Based on the sample, Is there enough evidence to reject the null? Use any method.

Answer 1

a) The parameter of interest is the true chance of prediction of the sex of an unborn child by women.

b) The null hypothesis is

where p is the chance of women making a correct prediction about the sex of an unborn child.

The alternate hypothesis is

c) The observed value of p is

d) Since the observed value is close to 0.5 and the sample size is large, so we can approximate the observed value to be distributed as normal.

The sample size is considered large when np and nq are both greater than 5.

Here n = 104 , p =0.5 , q = 1-p = 0.5

So, np = 104 * 0.5 = 52 which is greater than 5

and nq = 104 * (1-0.5) = 52 which is also greater than 5. Hence, the sample size is large.

e) The z statistics is given by

f) In order to reject the null, the value of should be greater than a critical value. That critical value, with a significance of 0.05 is given by

Hence, if is greater than this critical value of 0.581 then we will reject the null hypothesis.

g) Based on the sample, we have

Since the value of is less than the critical value which is 0.581 , hence we can't reject the null.

Thank You!!

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