Respuesta :
Answer:
We conclude that those celebrating Valentine's Day spend more than an average of $125 on gifts.
Step-by-step explanation:
We are given that Hallmark would like to test the hypothesis that those celebrating Valentine's Day will spend more than an average of $125 on gifts.
A random sample of 18 people celebrating Valentine's Day spent an average of $148.50 with a standard deviation of $34.90.
Let [tex]\mu[/tex] = average amount spent on gifts celebrating Valentine's Day
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\leq[/tex] $125 {means that those celebrating Valentine's Day spend less than or equal to an average of $125 on gifts}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > $125 {means that those celebrating Valentine's Day spend more than an average of $125 on gifts}
The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average amount spent = $148.50
s = sample standard deviation = $34.90
n = sample of people = 18
So, test statistics = [tex]\frac{148.50-125}{\frac{34.90}{\sqrt{18} } }[/tex] ~ [tex]t_1_7[/tex]
= 2.857
The value of the sample test statistics is 2.857.
Now, P-value of the test statistics is given by the following formula;
P-value = P( [tex]t_1_7[/tex] > 2.857) = 0.0056
Because in the t table the critical value of 2.857 at 17 degree of freedom will lie between P = 1% and P = 0.5%.
Now, since P-value of test statistics is less than the level of significance as 0.01 > 0.0056, so we sufficient evidence to reject our null hypothesis.
Therefore, we conclude that those celebrating Valentine's Day spend more than an average of $125 on gifts.
