Question

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims that its graduates will score higher, on average, than the mean score 600. A random sample of 70 students completed the course, and their mean SAT score in mathematics was 613. At the 0.05 level of significance, can we conclude that the preparation course does what it claims? Assume that the standard deviation of the scores of course graduates is also 48.Perform a one-tailed test. Then fill in the table below.Carry your intermediate computations to at least three decimal places, and round your responses as specified in the table.Null hypothesisAlternative hypothesisType of test statistic (if t, list the degrees of freedom)Value of test statisticCritical value at the 0.05 level significance (Round to at least three decimal places)Can we support the preparation course's claim that its graduate score higher in SAT? Yes or noI understand that this may seem like multiple questions, it is indeed however just one in terms of to conclude the final answer you must have these steps in mind.

Answer 1

Answer:

No

Step-by-step explanation:

Let's first start gathering and ordering the data provided by the problem

1. The mean SAT score is $\mu=600$, we are going to call it $\mu$ since it's the "true" mean

2. The standard deviation (we are going to call it $\sigma$) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$ and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$[tex]t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}= \frac{| 600-613 |}{48/\sqrt(70}}= \frac{| 13 |}{48/8.367}= \frac{| 13 |}{5.737}=2.266$[/tex]

since 2.266>1.645 we  can reject the null hypothesis.