According to a study on the effects of smoking by pregnant women on rates of asthma in theirchildren, for expectant mothers who smoke 20 cigarettes per day, 22.2% of their childrendeveloped asthma by the age of two in the US. A biology professor at a university would like totest if the percentage is less in another country. She randomly selects 320 women who onlydeliver one child and smoke 20 cigarettes per day during pregnancy in that country and findsthat 79 of the children developed asthma by the age of two. In this hypothesis test,Do the hypothesis test. Use a level of significance of a = 0.05;Use the unrounded values in Excel to find the answers for #3 and #4.hypothesis test.1. This is an example of a2. Find the standard error.3. Find the z value.4. Find the p-value.5. The biology professor should he null hypothesis. --

Population proportion of children who developed asthma p=22%=0.222The sample size for the study, n=320No. of children among " n " developed asthma, x=79The estimated proportion from the sample is, p^=79320=0.246875 The defined level of significance, α=0.05To determine:If the % of children who develop asthma is less in another country.1) The given hypothesis test is an example of a one-proportion z hypothesis test because it is focused on comparing the proportion of children developing asthma in a specific population to a known or hypothesized proportionBased on the given information and the objective of the study the null and alternative hypotheses can be formulated as,To test the above hypothesis z-test statistic is given by2 )The formula to compute the standard error is given by3 )The z-test statistic is given byComputing the Z value using Excel :4)The p-value can be formulated as Hence the p-value is 0.8578500785 )The decision rule in hypothesis testing using the p-value approach is as followsConclusion :1 ) This is an example of one proportion z hypothesis test.2 ) Standard error = 0.023232253 ) z = 1.070714 ) p-value = 0.8578500785 ) The null hypothesis is not rejected.