A: Calculate 92 percent confidence intervals for the average SAT score during 20

Question

A: Calculate 92 percent confidence intervals for the average SAT score during 2013. B: Some believe that the average SAT score was higher than 1010 in 2014, use single t-test to validate their claim (alpha: 5%) C: It seems there is no difference in average SAT score in 2013 Vs 2014. Use independent t-test to see if there was a significant difference in average SAT score in 2013 VS 2014 (alpha 10 %)

The attached data set contains information about students’ average SAT scores of 100 academic institutions during 2013 and 2014 Using this dataset, answer the following questions:

2013 2014 1 1302 1130 2 1154 1160 3 1240 1105 4 987 1171 5 1026 1053 6 879 1308 7 1069 990 8 1098 1134 9 1105 969 10 1045 1077 11 1001 1008 12 919 1217 13 924 1146 14 939 1145 15 853 1215 16 886 910 17 1095 877 18 993 1059 19 993 976 20 780 1116 21 755 1045 22 1080 1030 23 1025 1144 24 850 1065 25 1009 1104 26 1175 1055 27 1158 1281 28 1239 1081 29 1029 990 30 1123 940 31 1153 1150 32 1105 990 33 1035 1089 34 1080 918 35 1200 947 36 1110 802 37 1070 890 38 1192 1115 39 1028 1159 40 989 1101 41 859 1032 42 941 1050 43 1091 1030 44 1116 963 45 1231 918 46 1112 1089 47 1054 1292 48 974 1102 49 1102 1110 50 990 1085 51 1003 1300 52 1084 1050 53 1199 1024 54 1267 871 55 1009 945 56 1362 1030 57 837 1366 58 1068 945 59 892 870 60 1183 1048 61 1010 1030 62 1036 899 63 994 970 64 1038 1152 65 1032 1166 66 1018 946 67 1363 1002 68 930 1185 69 1017 1031 70 1035 1104 71 982 899 72 1192 946 73 1316 938 74 1072 988 75 957 925 76 1123 992 77 1161 1170 78 1050 974 79 1352 1029 80 946 1006 81 970 899 82 1391 1010 83 910 1004 84 971 1030 85 1050 1254 86 878 1287 87 980 1086 88 830 1296 89 990 1002 90 890 999 91 928 1066 92 1090 1070 93 1104 1005 94 1216 965 95 1036 1313 96 1057 1095 97 1067 1226 98 1208 1031 99 1310 1145 100 851 1054

Explanation / Answer

I am using R software to solve this problem.

At first we can load the data into R with read.table() command as below. I have copied the data into a text file and have given the column headers as "Institutes" "Data2013" "Data0214"

InputData <- read.table("Data.txt", header = T, sep = " ")

A) Here sample size = 100

Standard deviation of the 2013 data = sd(InputData$Data2013) = 133.1643

To calculate the confidence interval, we first need to calculate the margin of error.

ME = critical value * standard error

Here degrees fo freedom = Sample size - 1 = 100 - 1 = 99

Confidence level = 92%

So significance level = 1 - 0.92 = 0.08

Critical probability = 1 - 0.08/2 = 1 - 0.04 = 0.96

The critical value is the t statistic having 99 degrees of freedom and a cumulative probability equal to 0.96. The value ftom the t distribution is 1.769

SE = s / sqrt( n ) = 133.1643/sqrt(100) = 13.316

So margin of Error = Critical value * standard error

= 1.769 * 13.316 = 23.556

The range of the confidence interval is defined by the sample statistic + margin of error

Sample statistic = mean(InputData$Data2013) = 1054.18

So 92% confidence interval is (1054.18 - 23.556) to (1054.18 + 23.556) = 1030.624 to 1077.736

B) Here we need to examine if the average SAT score was higher than 1010 in 2014.

We can use the t.test() function in R for this. This is a one tailed test as we are interested in finding if average score is greater than 1010.

t.test(InputData$Data0214, mu = 1010, alternative = "greater")

   One Sample t-test

data: InputData$Data0214
t = 4.3164, df = 99, p-value = 1.882e-05
alternative hypothesis: true mean is greater than 1010
95 percent confidence interval:
1040.588 Inf
sample estimates:
mean of x
1059.71

Here we can see that the p value is very small, les than 0.05, so we can reject the null hypothesis. That means the average SAT score was indeed higher than 1010 in 2014.

C) Here we need to examine whether there is any difference in average SAT score in 2013 Vs 2014. Again we can use t.test() function to do independent t test. Here we need to use confidence level of 0.90.

t.test(InputData$Data2013, InputData$Data0214, alternative = "two.sided", conf.level = 0.90, var.equal = T)

   Two Sample t-test

data: InputData$Data2013 and InputData$Data0214
t = -0.31411, df = 198, p-value = 0.7538
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
-34.62468 23.56468
sample estimates:
mean of x mean of y
1054.18 1059.71

Here we can see that the p value is 0.7538, which is greater than 0.10. So there is no significant difference in average SAT score in 2013 Vs 2014.

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