00:01
Hello student, in the given question in part a, we have to determine the set of data, set of data supports the engineer's claim, supports the engineer's claim, engineer's claim that the mean moisture, mean moisture content is claim that mean moisture, moisture contain, content is not out of specification, is not out of specification.
00:55
So we can perform here one sample t -test, one sample t -test.
01:03
So first we have to formulate the null and alternative hypothesis as the null hypothesis is that the true mean moisture content, true mean moisture content is equals to 5 percent versus the alternative hypothesis h1 is true alternative hypothesis it is not equal to 5 percent means true mean moisture content, content, content sorry, content is not equals to 5 percent at alpha is equals to 0 .05.
02:13
So the formula for t -statistic is t is equals to, t is equals to x minus mu upon s upon under root of n which is equals to 5 .59 minus 5 upon 0 .5 upon under root of 10 which is equals to 3 .16.
02:44
So using t -distribution table and 9 degrees of freedom, the critical value, critical value is equals to 2 .262, 2 .262.
03:02
So since the absolute value of calculated t -statistic is greater than, t calculated is greater than critical value, critical value, therefore here we reject the null hypothesis, reject the null hypothesis, therefore the set of, therefore the set of data does not support, does not support the engineer's claim, the engineer's claim at the specified test level, at the specified test level, level...