An engineer suspects that the hardness readings for a particular piece of testing equipment differ depending on the type of tip he uses. he took ten pieces of metal and tested each using both tips. the data follow: • specimen 1 2 3 4 5 6 7 8 9 10 • tip 1 7 3 3 4 8 3 2 9 5 4 • tip 2 6 3 5 3 8 2 4 9 4 5 • perform the most appropriate hypothesis test using a significance level of 0.05. • construct the most appropriate 95% confidence interval. • what must you assume to do an analysis (hypothesis test or confidence interval) on these data? how comfortable are you with these assumptions?

- Because the value of the calculated value of t is less than the critical value (|t| < |tc| = -0.096 < 2.10), we can't reject the null hypothesis, which means that there is not enough evidence to conclude that the hardness readings of the metal pieces differ depending on the type of tip that it's used.

- Because the obtained confidence interval of -2.27 < μ1 - μ2 <  2.07 includes the null value (zero, 0), we can again assume that there is not a significant difference between the sample means of the two used tips, therefore, we can't reject the null hypothesis.

What must you assume to do an analysis (hypothesis test or confidence interval) on these data? We must assume that the sample was taken randomly, meaning that each value is sampled independently from each other value. We must assume also that the sample has a normal distribution and that the variances are equal.

How comfortable are you with these assumptions? very comfortable

Step-by-step explanation:

Specimen    1     2     3     4     5     6     7     8      9      10

Tip 1             7     3     3     4     8     3     2     9      5      4

Tip 2            6     3     5     3     8     2     4     9      4      5

Because the sample size is less than 30 (n = 10), and we don't actually know the standard deviation of the population, we will use the T-statistic to test the hypotheses.

H0: μ1 = μ2

H1: μ1 ≠ μ2

First we obtain the sample mean for each tip, and the sample standard deviation (SD):

X¯= ∑Xi/n

S = √(∑(X-X¯)²/n-1)

TIP 1                                     TIP 2

n1 = 10                                  n2 = 10

X1¯ = 4.8                              X2¯ = 4.9

S1 = 2.40                             S2 =  2.23

Then we need to calculate the pooled variance for the test:

σ² = (n1 - 1)S1² + (n2 - 1)S2² / n1 + n2 - 2

σ² = (10 -1)(2.40)² + (10 - 1)(2.23)² / 10 + 10 - 2 = 5.36

We proceed to calculate the T-statistic:

t = (X1¯ - X2¯) / √(σ²/n1 + σ²/n2)

t = (4.8 - 4.9) / √(5.36/10 + 5.36/10) = -0.096

Now we need to consult a two-tailed distribution T-table (observe attached image) to obtain the t critical value. For this we need to know that our degrees of freedom are calculated like n1 + n2 - 2, therefore df = 10+10-2=18; and we are using a significance level of 0.05, so with these two data we consult the T-table, and we obtain a value of: tc = 2.10

Because the value of the calculated value of t is less than the critical value (|t| < |tc| = -0.096 < 2.10), we can't reject the null hypothesis, which means that there is not enough evidence to conclude that the hardness readings of the metal pieces differ depending on the type of tip that it's used.

To construct the confidence interval for this case we will use the next equation:

(X1¯ - X2¯) - (tα/2,df) Sp√(1/n1 + 1/n2) < μ1 - μ2 < (X1¯ - X2¯) + (tα/2,df) Sp√(1/n1 + 1/n2)

To obtain the pooled standard deviation we just calculate the square root of the already calculated pooled variance (σ²), therefore:

Sp = √σ² = √5.36 = 2.31

As the confidence level is of 95%, the significance level is α=0.05, therefore:

tα/2,df = t0.025,18

If we look for the t value in a one-tailed distribution T-table (observe attached image), for that significance level and those degrees of freedom, we obtain:

t0.025,18 = 2.10

Now we calculate the confidence interval:

(4.8 - 4.9) - (2.10) 2.31√(1/10 + 1/10) < μ1 - μ2 < (4.8 - 4.9) + (2.10) 2.31√(1/10 + 1/10) = -2.27 < μ1 - μ2 <  2.07

Because the obtained confidence interval of -2.27 < μ1 - μ2 <  2.07 includes the null value (zero, 0), we can again assume that there is not a significant difference between the sample means of the two used tips, therefore, we can't reject the null hypothesis.

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step-by-step explanation: