A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawnequipment manufacturer, located in Omaha, feels the estimate is too low for households in Omaha. Test theclaim that the proportion with lawn mowers in Omaha is higher than 65%. Among 497 randomly selectedhomes in Omaha, 340 had one or more lawn mowers.

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

d 600 because 15*10*2=300

300+300=600

step-by-step explanation:

45.36 feet per second

step-by-step explanation:

tyrone travels 30 miles in 58.2 minutes.  

1 mile = 5280 feet; this means he travels

30(5280) = 158400 feet in 58.2 minutes.

1 minute = 60 seconds; this means he travels 158400 feet in

58.2(60) = 3492 seconds.

this gives us the unit rate of

158400/3492 = 45.36 feet per second.