Question

Suppose Heather suspects that the B string on her guitar is too sharp. She knows that playing a B3 note should produce a frequency of 246.94 Hz, and if the frequency is higher than that, the note is sharp. Because of slight variations in her guitar tuner's measurements, she played a B3 on the B string 10 times and recorded each of the measured frequencies.

Heather found that her 10 measurements yielded a sample mean of 247.09 Hz and a sample standard deviation of 0.39 Hz. She then used this information to conduct a one-sample ?‑test of the null hypothesis ?0:?=246.94 against the alternative hypothesis ?1:?>246.94, where ? is the mean frequency, in hertz, produced when she plays a B3 note before tuning her guitar.

Heather calculated a test statistic of 1.22 and a ?-value of 0.1274. If she is conducting her test at a significance level of 0.05, what is her conclusion?

The decision is to fail to reject the null hypothesis at significance level ?=0.05 (?=__)

There is insufficient evidence that the mean frequency for all B3 notes played before tuning is more than __ Hz

What are p and Hz?

Answer 1

Date : Ho :4=246.94 mkon - 247.09 0.39 t = x-Mo This is a problem of one sample t-test. Null hypothesis Alternate hypothesis He: 47246.94 This is a right-tailed t-test for one mean with unknown population standard deviation. So, test statistic X= Sample 8. Sample so Sva n = Sample fize=10 - 247.09 - 346.94 Mo= 246.94 0.39 / Vió 1:22 (approx) And it is given, p-value = 0.1274, for t-1-22 Significond level 2-0.05 So, according to the above figures, we find that, the p-value is p= 0.1274 and it is greater than the significance lever X= 0.05 p=0.1274 7,0:05 22 so we can't reject the null hypothests. Thus it is concluded that © The decision is to fail to reject the null hypothesis at significant level 20.05. because the p-value for test statistics to 1.22 is given greater than Significant level 220.05. O it is also concluded that > There is insufficient evidence that the mean frequency for all B3 notes played before turning, is more than 246.94 Hz. Because we failed to reject hull hypothesis. p = p value for t= 1:22' and 220.05' and 'ne 10' Hz frequency for null hypothesis 246.94 Hz. p=0.1274 2 0.1234 Scanned by TapScanner