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tamaranim1 [39]
2 years ago
8

SOMEONE HELP!!!!!! Don’t use what websites! And please like list answer 1,2,3, and 4

Mathematics
1 answer:
harkovskaia [24]2 years ago
8 0

Answer:

1. 24  2. 20  3. 1.75(I think)  4. 11

Step-by-step explanation:

3 might be incorrect but The others seem more correct than ever :)

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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

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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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3 years ago
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