KA simple random sample of pulse rates of 30 women from a normally distributed population results in a standard deviation of 12.5 beats per minute. The normalrange of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standarddeviation of 10 beats per minute. Use the sample results with a 0.05 significance level to test the claim that pulse rates of women have a standard deviationequal to 10 beats per minute. Complete parts (a) through (d) below.b. Compute the test statistic.x2²=0(Round to three decimal places as needed.)c. Find the P-value of the test statistic.The P-value of the test statistic is(Round to three decimal places as needed.)K

The information provided in the question are as : Sample size (n) = 30, Sample standard deviation (s) = 12.5 Population standard deviation (σ) = 10 Null and alternative hypotheses : Null hypothesis — Pulse rate of women is 10 beats per minute or the standard deviation is equal to 10 beats per minute . Symbolically, H0 : σ = 10 Alternative hypothesis — The standard deviation is not equal to 10 beats per minute i.e., Ha : σ ≠10Part (b) : The chi-square test statistic formula is given by, χ2 = ((n–1) × s2)/σ2 On substituting the values in the above formula we get, χ2 = ((30–1) × 12.52)/102 χ2 = (29 × 156.25)/100 χ2 = 4531.25/100 χ2 = 45.3125 Rounding to three decimal places we get, χ2 = 45.313 (ANSWER) Part (c) : The degree of freedom for chi-square test statistic is calculated as : Degree of freedom = n–1 Degree of freedom = 30–1 = 29 Since, our test is two-tailed so the p-value is calculated as, p-value = 2 × p-value (RT) Where, p-value (RT) = p-value for right tail. As the chi-square distribution table cannot be used to find the exact p-value. Hence, using R function to calculate the p-value. R - function : pchisq(q, df, lower.tail = FALSE ) Where, q = test statistic, df = degree of freedom On providing the input : pchisq(45.313, 29, lower.tail = FALSE ) We get the output, p-value (RT) = 0.02742731 The p-value of chi-square test-statistic for 29 degree of freedom is given by, p-value = 2 × 0.02742731 p-value = 0.05485462 Rounding to three decimal places we get, p-value = 0.055 ANSWER : p-value = 0.055Conclusion : Fail to reject the null hypothesis if p-value > significance level. Reject the null if p-value ≤ significance level. Since, p-value = 0.055 is greater than the level of significance = 0.05. Hence, we fail to reject the null hypothesis.