Respuesta :
Answer:
a)H0: m = 700
H1: m < 700
b) This is a left-tailed test because our alternative hypothesis is m<700 and the critic region is going to be the left tail of the critic point.
c) z = -1.22
d) [tex]p-value=P(z<-1.22)=0.1112[/tex]
e) We fail to reject the null hypothesis
f) There isn't evidence to said that the population mean number of such meals is less than 700.
Step-by-step explanation:
We want to know if the population mean is less than 700, so we define the null and alternative hypothesis as:
H0: m = 700
H1: m < 700
Where m is the mean number of meals. This is a left-tailed test because our alternative hypothesis is m<700 and the critic region is going to be the left tail of the critic point.
Then, taking into account that the size of the sample is bigger than 30 and we know that the population standard deviation is equal to 49, the appropriate statistic is:
[tex]z=\frac{x-m}{s/\sqrt{n} }[/tex]
Where x is the mean of the sample, s is the population standard deviation, n is the size of the sample and m is 700, .
So, replacing x by 690, s by 49 and n by 36, we get that the test statistic is equal to:
[tex]z=\frac{690-700}{49/\sqrt{36} }=-1.22[/tex]
Therefore, using the standard normal distribution table, the p-value is equal to:
[tex]p-value=P(z<-1.22)=0.1112[/tex]
Finally, if we compare the p-value with the level of significance we can conclude that we fail to reject the null hypothesis (p-value>0.01).
It means that with a significance level of 0.01, there isn't evidence to said that the population mean number of such meals is less than 700.
