Question

A hospital takes record of any birth that occurs there every day. On one day, the hospital reports that 35 of the 62 babies born were girls. Assuming that all of the parents did not have any gender selection procedures, there is a probability of 0.31 of getting these results by chance. Do these results have statistical significance at the 0.05 level of significance?

Answer 1

Claim: The result have statistical significance. That is $p \neq 0.31$

The null and alternative hypothesis are:

H₀: There is no statistical significance that is p = 0.31

H₁: There is statistical significance that is $p \neq 0.31$

Test statistics:

The formula of z-test statistics is,

$z = \frac{\hat p - p}{\sqrt{\frac{p(1-p)}{n}}}$

Where $\hat p - sample \ proportion$

n = total number of babies born = 62

x = number of baby girls born = 35

$\hat p =\frac{x}{n} = \frac{35}{62} = 0.564516129$

$z = \frac{\hat p - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.564516129-0.31}{\sqrt{\frac{0.31(1-0.31)}{62}}} = 4.33$

Critical value:

Here the alternative hypothesis contains not equal to sign, so the test is a two-tailed test.

There are two critical values, both are the same just opposite in sign.

alpha = 0.05 (level of significance)

Alpha/2 = 0.05/2 = 0.025

Using z table the critical value corresponding to 0.025 is -1.96, that is occurs at -1.9th row and 0.06th column.

So the critical values are -1.96 and 1.96

Decision rule:

If the z-test statistics falls between critical values then fail to reject the null hypothesis otherwise reject the null hypothesis.

Here z test statistics 4.33, does not fall between -1.96 and 1.96 so reject the null hypothesis.

Conclusion: Reject the null hypothesis, that is there is statistical significance at the 0.05 level of significance.

Answer 2

No, Less thn 0.05

Answer 3

no, greater than 0.05