Question

In: Statistics and Probability

A fertility clinic claims a 40% success rate for in-vitro fertilization (IVF). In a study involving...

A fertility clinic claims a 40% success rate for in-vitro fertilization (IVF). In a study involving 120 women who initiated IVF at the clinic, 39 live births. Is there evidence to reject the clinic’s claim at the 5% level of significance?

Expert Solution

Solution :

Given that,

= 40% = 0.40

1 - = 1 - 0.40 = 0.60

n = 120

x = 39

Point estimate = sample proportion = = x / n = 39 / 120 = 0.325

This is a two tailed test.

The null and alternative hypothesis is,

Ho: p 0.40

Ha: p 0.40

The test statistics,

z = ( - ) / (*(1 - ) ) / n

= ( 0.325 - 0.40) / (0.40*0.60) / 120

= -1.677

P-value = 2* P(Z < z)

= 2*P(Z < -1.677)

= 2*0.0468

= 0.0936

The p-value is p = 0.0936, and since p = 0.0936 ≥ 0.05, it is concluded that the null hypothesis is fails to reject.

Conclusion:

It is concluded that the null hypothesis Ho is fails to reject. Therefore, there is not enough evidence to claim that the population proportion p is different than , at the α = 0.05 significance level.

orchestra answered 2 years ago

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