A diet center claims that it has the most effective weight loss program in the region. Its advertisements say, 'Participants in our program lose more than 7 pounds within a month'. A sample of 10 clients of this program are weighed on the first day of the diet and then one month later.
Client Day1 One Month Later
1 185 179
2 163 217
3 167 212
4 205 130
5 182 181
6 165 158
7 212 202
8 130 124
9 217 211
10 181 172
1. What is the p-value of the test (round to 4 decimal places)?

P-value = 0.3334

Step-by-step explanation:

We have a hypothesis test, with this null and alternative hypothesis:

(The null hypothesis states that the weight loss is equal or higher than 7)

The significance level is assumed as 0.05.

From the table, we have a sample mean of 2.1 and a sample standard deviation of 34.7.

The t-statistic is

For 9 degrees of freedom, the P-value for t=-0.445 is P(t<-0.445)=0.3334.

This P-value is bigger than the significance level, so the effect is not significant. The null hypothesis is not rejected.

answer: no picture is

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;