Respuesta :
Answer:
P-value = 0.3334
Step-by-step explanation:
We have a hypothesis test, with this null and alternative hypothesis:
[tex]H_0: \mu\geq7\\\\H_a:\mu<7[/tex]
(The null hypothesis states that the weight loss is equal or higher than 7)
The significance level is assumed as 0.05.
From the table, we have a sample mean of 2.1 and a sample standard deviation of 34.7.
The t-statistic is
[tex]t=\frac{M-\mu}{\sigma/\sqrt{n}} =\frac{2.1-7}{34.7/\sqrt{10}} =\frac{-4.9}{11}= -0.445[/tex]
For 9 degrees of freedom, the P-value for t=-0.445 is P(t<-0.445)=0.3334.
This P-value is bigger than the significance level, so the effect is not significant. The null hypothesis is not rejected.
Answer:
The p-value is 0.4264
Step-by-step explanation:
In calculating the P-value of the test, the standard deviation and mean differences of the weight on first day of diet and weight one month later is computed.
Arranging the data on a table, we have;
s/no Weight on First Weight on M = F-L M² (M-Mbar)²
Day of Diet (F) Month (L)
1 185 179 6 36 15.21
2 163 217 -54 2916 3147.21
3 167 212 -45 2025 2218.41
4 205 130 75 5625 5314.41
5 182 181 1 1 1.21
6 165 158 7 49 24.01
7 212 202 10 100 62.41
8 130 124 6 36 15.21
9 217 211 6 36 15.21
10 181 172 9 81 47.61
TOTAL: 21 10735 10860.9
Mean ( M bar) = Σ M/n where n = 10
= 21/10
= 2.1
Standard deviation (sd) = √Σ( M - M bar)²/n-1
= √10860.9/10-1
= √10860.9/9
= √1206.767
= 34.74
Calculating the t-test value, we use the formula;
T-test = Mbar/(sd/√n)
= 2.1/(34.74/√10
= 2.1/ (34.74/3.16)
= 2.1 /10.994
= 0.1910
From t-test table, the p-value for 9 degree of freedom (n-1) is given as 0.4264
