An experiment was performed to compare the wear of two different laminated materials. Twelve pieces of material 1 were tested by exposing each piece to a machine measuring wear. Ten pieces of material 2 were similarly tested. In each case, the depth of wear was observed. The samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4, while the samples of material 2 gave an average of 81 with a sample standard deviation of 5. Can we conclude at the 0.05 level of significance that the abrasive wear of material 1 exceeds that of material 2 by more than two units? Assume the populations to be approximately normal with equal variances pdf

At 0.05 level of significance, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Step-by-step explanation:

We hypothesize that mean difference between abrasive wear of material 1 and material 2 is greater than 2.

So we write the null hypothesis ,

and the alternative hypothesis .

We will find the T-score as well as the p-value. If the p-value is less than the level of significance, we will reject the null hypothesis, i.e. we will conclude that the abrasive wear of material 1 is less than that of material 2. Otherwise, we will accept the null hypothesis.

Since the variance is unknown and assumed to be equal, we will use the pooled variance

,

where .

The mean of material 1 and material 2 are respectively and mean difference is equal to 4. The hypothesize difference is equal to 2.

To find the T-score, we use the following formula

Substituting all the values into the T-score formula gives us , and the respective p-value is equal to 0.31. This means we have enough statistical evidence not to reject the null hypothesis, and at 5% significance level, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

which expression can be used to convert 80 us dollars (usd) to australian dollars (aud)?

1 usd = 1.0343 aud

1 aud = 0.9668 usd

step-by-step explanation:

sorry just need the points