Respuesta :
Answer:
a) The null and alternative hypothesis are:
[tex]H_0: \mu=13.04\\\\H_a:\mu< 13.04[/tex]
b) P-value = 0.0151
c) We reject the null hypothesis.
We conclude that the cost of a restaurant meal is significantly cheaper than a comparable meal fixed at home.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the mean cost of a restaurant meal is significantly less than fixing a comparable meal at home.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=13.04\\\\H_a:\mu< 13.04[/tex]
The significance level is 0.05.
The sample has a size n=100.
The sample mean is M=12.6.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2}{\sqrt{100}}=0.2[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{12.6-13.04}{0.2}=\dfrac{-0.44}{0.2}=-2.2[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=100-1=99[/tex]
This test is a left-tailed test, with 99 degrees of freedom and t=-2.2, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t<-2.2)=0.0151\\[/tex]
As the P-value (0.0151) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the mean cost of a restaurant meal is significantly less than fixing a comparable meal at home.
There is enough evidence to support the claim that the mean cost of a restaurant meal is significantly less than fixing a comparable meal at home.
The hypotheses
The null hypothesis is always represented by the equality sign.
So, the null hypothesis is:
[tex]H_o : \mu = 13.04[/tex]
For the meal to be cheaper, then the less than inequality sign would be used.
So, the alternate hypothesis is:
[tex]H_a : \mu < 13.04[/tex]
Hence, the null and the alternate hypotheses are:
[tex]H_o : \mu = 13.04[/tex] and [tex]H_a : \mu < 13.04[/tex]
The p value for a sample size of 100
The given parameters are:
n = 100
[tex]\sigma = 2[/tex]
[tex]\bar x = 13.04[/tex]
[tex]\mu = 12.60[/tex]
Start by calculating the standard error
[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]
This gives
[tex]\sigma_x = \frac{2}{\sqrt {100}}[/tex]
[tex]\sigma_x = 0.2[/tex]
Next, calculate the test statistic
[tex]t = \frac{\mu - \bar x}{\sigma_x}[/tex]
This gives
[tex]t = \frac{12.60 - 13.04}{0.2}[/tex]
[tex]t = -2.2[/tex]
Calculate the degrees of freedom
[tex]df = n - 1[/tex]
[tex]df = 100 - 1[/tex]
[tex]df = 99[/tex]
Next, we calculate the p-value as follows:
[tex]p = P(t < -2.2)[/tex]
At a degree of freedom of 99, the p value when the t value is less than -2.2 is:
[tex]p = 0.0151[/tex]
Hence, the p value for a sample size of 100
The conclusion
We have:
The significance level, α = 0.05
The p-value in b, 0.0151 is lesser than the significance level, 0.05.
This means that we reject the null hypothesis
So, the conclusion is:
There is enough evidence to support the claim that the mean cost of a restaurant meal is significantly less than fixing a comparable meal at home.
Read more about hypothesis testing at:
https://brainly.com/question/15980493
