A random sample of 830 births included 425 boys use a 010...

A random sample of 830 births included 425 boys. Use a 0.10 significance level to test the claim that 50.6% of newborn babies are boys. Do the results support the belief that 50.6% of newborn babies are boys?

Transcript

00:01 In this question, we have been given to test the claim based on the given information.

00:05 So, here the sample size n will be equals to 830 and the number of favorable cases is here x and that is equals to 425 and the value of sample proportion will be here x divided by n and x value is here 425 and then n value is here 830.

00:31 Therefore, the value of p hat will be here 0 .512 and hence we can say that the significance level will be here 0 .10.

00:43 Now solving this very problem further, it can be said that in the first part we are having the null hypothesis as p is equals to 0 .506 versus the alternate hypothesis which can be written such that p is not at all equals to 0 .506.

01:05 So, this corresponds to two -tailed test.

01:08 So, it will be here tailed and now here it is test for which a z -test for one population proportion needs to be used and hence here in the second part we can say that the rejection reason can be written based on the information provided that the significance level is 0 .10 and the critical value for a two -tailed test will be here equals to 1 .64 and because of this the rejection reason let us here write it will be rejection and then here it is reason.

01:46 So, this very value is here going to come around we can say r and it can be denoted as it is equals to z such that this semicolon means sorry this colon means such that here it will be modulus of z greater than 1 .64 and this is basically the rejection reason.

02:09 Now in the c part of this very question we can say that we have to calculate the value of z that is test statistic that is p bar minus p naught divided by it will be here under root of we can say p naught multiplied with 1 minus p naught and then in the denominator it will be n.

02:30 So, that is here equals to 0 .512 minus 0 .506 and then in the denominator it will be under root of let us write it is 0 .506 and then here it is multiplied with 1 minus 0 .506 and then in the denominator it will be 830.

02:54 So, the value of z will be here 0 .349 and we have obtained the value of test statistic.

03:04 So, in the fourth part we will be writing the decision...

Question

A random sample of 830 births included 425 boys. Use a 0.10 significance level to test the claim that 50.6% of newborn babies are boys. Do the results support the belief that 50.6​% of newborn babies are boys?

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A random sample of 830 births included 425 boys. Use a 0.10 significance level to test the claim that 50.6% of newborn babies are boys. Do the results support the belief that 50.6% of newborn babies are boys?