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ella [17]
3 years ago
9

Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

of 1085 and standard deviation of 52. Grade 2 has 15 observations yielding a sample mean of 1034 and standard deviation of 61. Assume the populations to be approximately normal with equal variances.
State and conclude your hypothesis at the 0.05 level of significance if grade 1 and grade 2 true means are equal each other?

Find P-Value.
Mathematics
1 answer:
LuckyWell [14K]3 years ago
8 0

Answer:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

df = n_1 +n_2 -2 = 10+15-2= 23

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

\bar X_1 = 1085 sample mean for group 1

\bar X_2 = 1034 sample mean for group 2

n_1 = 10 sample size for group 1

n_2 = 15 sample size for group 2

s_1 = 52 sample deviation for group 1

s_2 = 61 sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: \mu_1= \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

And the statistic is given by:

t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}

And replacing we got:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 10+15-2= 23

And the p value would be:

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

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