A private college advertised that last year their freshman students, on average, had a score of 1120 on the college entrance exa
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Answer:
Step-by-step explanation:
Given that:
sample size n = 36
standard deviation = 10.1
level of significance ∝ = 0.10
The null hypothesis and the alternative hypothesis can be computed as follows:
The test statistics can be computed as follows:
degree of freedom = n - 1 = 36 - 1 = 35
Since this test is two tailed .
The P -value can be determined by using the EXCEL FUNCTION ( = 2 × CHIDIST(35.7035, 35)
P - value = 2 × 0.435163515
P - value = 0.8703 ( to four decimal places)
Decision Rule : To reject the null hypothesis if P - value is less than the 0.10
Conclusion: We fail to reject null hypothesis ( accept null hypothesis) since p-value is greater than 0.10 and we conclude that there is sufficient claim that the normal range of pulse rates of adults given as 60 to 100 beats per minute resulted to a standard deviation of 10 beats per minute.
