At 5% there is significant evidence to reject the null hypothesis. You can conclude that there is a difference between the population mean of the wrinkle recovery angle of fabric treated with Permafresh and the population mean of the wrinkle recovery angle of fabric treated with Hylite.

X[bar]₁= 120.8 degrees

X[bar]₂= 162.75 degrees

S₁²= 14.5155 degrees

S₂²= 24.4727 degrees

Step-by-step explanation:

Hello!

The study variable is X: wrinkle recovery angle for a fabric specimen.

There is a suspicion that there is a difference in recovery from wrinkles after washing between two products (Permafresh and Hilite). To test this suspicion two random samples of fabric, 5 were treated with Permafresh and 4 were treated with Hylite resulting in the data:

Sample 1 (Permafresh)

X₁: wrinkle recovery angle for a fabric specimen treated with Permafresh.

X₁~N(μ₁;σ₁²)

n₁= 5

124; 104; 142; 111; 123

Sample mean X[bar]₁= (∑x₁i)/n₁= 604/5= 120.8 degrees

Sample variance S₁²= $\frac{1}{(n₁-1)}$ [(∑x₁²i)-(∑x₁i)²/n₁] = $\frac{1}{4}$ [(73806)-(604)²/5]= 210.7 degrees²

S₁= 14.515 ≅ 14.52 degrees

Sample 2 (Hylite)

X₂: wrinkle recovery angre for a fabric specimen trated with Hylite.

X₂~N(μ₂;σ₂²)

n₂= 4

147; 199; 149; 156

Sample mean X[bar]₂= (∑x₂i)/n₂= 651/4= 162.75 degrees

Sample variance S₂²= $\frac{1}{(n₂-1)}$ [(∑x₂²i)-(∑x₂i)²/n₂] = $\frac{1}{3}$ [(107747)-(651)²/4]= 598.916 ≅ 598.92 degrees²

S₂= 24.472≅ 24.47 degrees

To test the suspicion that there is a difference between the winkle recovery angle on fabric samples treated with Permafresh and Hylite, the hypothesis is:

H₀: μ₁ = μ₂

H₁: μ₁ ≠ μ₂

α: 0.05

To test the difference between the population means, considering that only sample information is available and the size of both samples, the most appropriate statistic to use is a pooled t for independent samples (unknown but equal population variances):

t= (X[bar]₁ - X[bar]₂) - (μ₁ - μ₂) ~t $_{n_1+n_2-2}$

Sa√(1/n₁+1/n₂)

Sa²= (n₁-1)S₁² + (n₂-1)S₂² = (4*210.7)+(3*598.92) = 377.08

n₁ + n₂ - 2 5+4-2

Sa= 19.418≅ 19.42

t= (X[bar]₁ - X[bar]₂) - (μ₁ - μ₂) = (120.8-162.75) - 0 = -3.22

Sa√(1/n₁+1/n₂) 19.42√(1/5+1/4)

Using the critical region approach, this rejection region is two-tailed with critical values:

$t_{n_1+n_2-2; \alpha/2 } = t_{7; 0.025 } = -2.365$

$t_{n_1+n_2-2; 1- \alpha/2 } = t_{7; 0.975 } = 2.365$

If t ≤ -2.365 or t ≥2.365, then the decision is to reject the null hypothesis.

If -2.365 < t < 2.365, then the decision is to not reject the null hypothesis.

Since the calculated t-value is less than the left critical value, the decision is to reject the null hypothesis.

I've calculated the p-value for the test: 0.0146

This p-value is less than the significance level of 0.05, then, using this approach, the decision is also to reject the null hypothesis.

This means, that at 5% there is significant evidence to reject the null hypothesis. You can conclude that there is a difference between the population mean of the wrinkle recovery angle of fabric treated with Permafresh and the population mean of the wrinkle recovery angle of fabric treated with Hylite.

I hope it helps!

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4 years ago