It looks like the terms are:
(7x2 + 5) and (x - 3)
First we need to write the null and alternate hypothesis for this case.
Let x be the average number of text message sent. Then
Null hypothesis: x = 100
Alternate hypothesis: x > 100
The p value is 0.0853
If p value > significance level, then the null hypothesis is not rejected. If p value < significance level, then the null hypothesis is rejected.
If significance level is 10%(0.10), the p value will be less than 0.10 and we reject the null hypothesis and CAN conclude that:
The mean number of text messages sent yesterday was greater than 100.
If significance level is 5%(0.05), the p value will be greater than 0.05 and we cannot reject the null hypothesis and CANNOT conclude that:
The mean number of text messages sent yesterday was greater than 100.
Answer:
Kindly check explanation
Step-by-step explanation:
Given :
Sample size, n = 30
Tcritical value = 2.045
Null hypothesis :
H0: μ = 9.08
Alternative hypothesis :
H1: μ≠ 9.08
Sample mean, m = 8.25
Samole standard deviation, s = 1.67
Test statistic : (m - μ) ÷ s/sqrt(n)
Test statistic : (8.25 - 9.08) ÷ 1.67/sqrt(30)
Test statistic : - 0.83 ÷ 0.3048988
Test statistic : - 2.722
Tstatistic = - 2.722
Decision region :
Reject Null ; if
Tstatistic < Tcritical
Tcritical : - 2.045
-2.722 < - 2.045 ; We reject the Null
Using the α - level (confidence interval) 0.05
The Pvalue for the data from Tstatistic calculator:
df = n - 1 =. 30 - 1 = 29
Pvalue = 0.0108
Reject H0 if :
Pvalue < α
0.0108 < 0.05 ; Hence, we reject the Null
