Question

 

1. On the following page are the exam scores on the first Statistics test for all my classes. Using everything we covered in the first three chapters of our textbook, describe the data. I recommend going through your notes and textbook, chapter by chapter. Include as much as you can – type of data, frequency distribution, histogram, numerical methods, etc. The standard deviation for the data is 16.7.

For problems 2 and 3, identify the null and alternative hypotheses, test statistic, critical value(s) and critical region or p-value, as indicated, and state the final conclusion that addresses the problem. Show all seven steps.

2. Every semester, I would like for more than 75% of my students to score a 70 or higher on the first test. This semester, out of the 72 students who took the first test, 59 got at least a C (scored higher a 70 or higher). Is there sufficient evidence to conclude, at the 10% significance level, that more than 75% of the students got at least a C on the first exam? Find the p-value.

3. A manufacturer makes ball bearings that are supposed to have a mean weight of 30 g. A retailer suspects that the mean weight is not 30 g. The mean weight for a random sample of 16b ball bearings is 28.4 g with a standard deviation of 4.5 g. At the 0.05 significance level, test the claim that the sample comes from a population with a mean not equal to 30 g. Find the critical value(s) and critical region.

 

4. The Precision Scientific Instrument Company manufacturers thermometers that are supposed to give readings of 0°C at the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing point of water, some give readings below 0°C and some give readings above 0°C. Assume that the mean reading is 0°C and the standard deviation of the readings is 1.00°C. Also assume that the frequency distribution of errors closely resembles the normal distribution. A thermometer is randomly selected and tested. If 10.2% of the thermometers are rejected because they have readings that are too high and another 10.2% are rejected because they have readings that are too low, find the two readings that are cutoff values separating the rejected thermometers from the others.

Exam Scores on the First Statistics Test

100	88
100	86
100	86
100	86
100	85
98	84
98	83
98	82
98	81
98	81
97	79
97	76
97	76
97	75
97	75
97	75
96	74
95	74
95	73
94	68
94	66
94	64
94	64
94	56
93	56
93	52
92	50
92	45
92	45
92	43
92	38
92	38
92
91
91
90
89
88
88
88

Answer 1

1. It is given that 58 students in the class scored more than 70 to get a C grade. Therefore the observed proportion of students who scored C =58/72 = 0.8056. The teacher asserts," I would like for more than 75% of my students to score a 70 or higher on the first test.

Null Hypothesis: The proportion who scored more than 70% =75% or

where in our case.

Alternative Hypothesis: The proportion who scored a C is more than 75% or 0.75

Test Statistic:

follows a Normal distribution.

The critical value of Z is 1.6448. The p-value corresponding to 1.0876 is NORM.S.DIST(-1.0875,TRUE)=0.1384. Since the p value is more than 0.05, the null hypothesis is not rejected.  

2. the mean weight of a ball bearing is supposed to be 30 g. In the sample of 16 ball bearings weigh 28.4 gm with SD 4.5 g. We need to test the claim that t " the sample comes from a population with a mean not equal to 30 g.". We are given that

.  

Assumptions: We shall assume that the weights of ball bearing is independent and follow a normal distribution.

Null Hypothesis: The population mean of the weight of ball bearings is 30 g.

Alternate Hypothesis: The population mean is not equal to 30 g

Level of significance :

Test Statistic:

has a distribution .

For this the critical value is 2.1314 and the p-value is 0.1754. Since the p-value is >0.05, we do not reject the null hypothesis.

3. It is given that the thermometer readings has a mean of 0 and the large sample variance(SD) is 1. Ie it is a standard normal variate. It is given that . Therefore the value of is 1.2702. Hence, the cutoff values are degree celsius and the upper cutoff value is 1.2702 degree celsius.