the total wages Jimmy makes at a local ice cream shop vary directly with the number of hours he works . He makes $8.50 per hour.
2 answers:
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Answer:
If we compare the p value obtained with the significance level assumed we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65
Step-by-step explanation:
Data given and notation
n=497 represent the random sample taken
X=340 represent the homes in Omaha with one or more lawn mowers
estimated proportion of homes in Omaha with one or more lawn mowers
is the value that we want to test
represent the significance level
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the true proportion of homes in Omaha with one or more lawn mowers is higher than 0.65.:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
If we compare the p value obtained with the significance level assumed we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65