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ch4aika [34]
3 years ago
13

Hi sorry can someone remove this question? it was a mistake. I dont know how to remove it...

Mathematics
1 answer:
N76 [4]3 years ago
5 0
Sorry I’m not sure either I also tried to remove one of mine.
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Read 2 more answers
Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean
LuckyWell [14K]

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

8 0
4 years ago
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