The chamber of commerce of a Florida Gulf Coast community advertises that area residential
property is available at a mean cost of $125,000 or less per lot. Suppose a sample of 32
properties provided a sample mean of $130,000 per lot and a sample standard deviation of
$12,500. Use a .05 level of significance to test the validity of the advertising claim

No, the advertising claim was not true.

Step-by-step explanation:

We are given that the chamber of commerce of a Florida Gulf Coast community advertises that area residential  property is available at a mean cost of $125,000 or less per lot. For this a random sample of 32  properties provided a sample mean of $130,000 per lot and a sample standard deviation of  $12,500.

We have to test that the validity of the advertising claim.

Let, Null Hypothesis, : $125,000 {means that the advertising claim is true}

Alternate Hypothesis, : > 5200 {means that the advertising claim is not true}

The test statistics that will be used here is One-sample t-test ;

             T.S. = ~

where, Xbar = sample mean = $130,000 per lot

              s = sample standard deviation = $12,500

              n = sample of properties = 32

So, test statistics = ~

                            = 2.263

Now, at 0.05 significance level the critical value of t at 31 degree of freedom in t table is given as 1.6955. Since our test statistics is more than the critical value of t which means our test statistics will lie in the rejection region. So, we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the area residential  property is not available at a mean cost of $125,000 or less per lot which means the advertising claim was not true.

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !