A diet plan claims that women have a mean weight of 145 lbs. Assume that the weights of women have a standard deviation of 30.86 lb. A sample of 40 weights of women has a mean weight of 153.2 lb. Find the P-value and, using a 0.05 significance level, state the conclusion about the null hypothesis. 0.0465; reject the null hypothesis 0.0930; fail to reject the null hypothesis

0.0930; fail to reject the null hypothesis

Step-by-step explanation:

Sample Statistics

Population Statistics

Claim

The mean weight of women is 145lbs

Counterclaim:

The mean weight of women is not equal to 145lbs

Null hypothesis

Alternative hypothesis

Significance Level

for a 2-tailed test at a 95% confidence level

Test Statistic

Corresponding p-value

The test is 2-tailed because our alternate hypothesis, , already implies that the population mean, , could be either less than or greater than 145lb. Since 0.09286>0.05, we are within the 95% confidence level, so there's insufficient evidence in our sample data to suggest that we reject the null hypothesis. This means that it's 9.3% more likely that the null hypothesis is true than the alternate hypothesis is. Thus, the sample data do not differ significantly from the expected mean of 145lb.

The answer is in the image 

The answer is in the image 

The total nom of  code that can be used is equal to 5+3 = 8

The answer is in the image 

Salesperson will make 6% of 1800

=(6/100)*1800

=108

Salesperson will make $108 in $1800 sales

The answer is in the image 

The solution is given in the image below

Salesperson will make 6% of 1800

=(6/100)*1800

=108

Salesperson will make $108 in $1800 sales

The answer is in the image