In: Statistics and Probability
If Belson Electronic's new converter produces less than the 110 volts of their old product, they must redesign it. A sample of 303 converters generated a standard deviation of 47.3 volts. Set Alpha at 5 percent. a mean of 107.2 volts, with State and test the proper set of hypotheses. Is a redesign necessary?
If Belson Electronic's new converter produces less than the 110 volts of their old product, they must redesign it. A sample of 303 converters generated a standard deviation of 47.3 volts. Set Alpha at 5 percent. a mean of 107.2 volts, with State and test the proper set of hypotheses. Is a redesign necessary?
Let the population mean voltage generated by a new converter be μ.
Step 1
H0: μ > 110 volts (counter claim)
H1: μ < 110 volts (claim)
This is a one-tailed(or left-tailed), one sample t test.
Step 2
Write down the data you have,
The sample mean (x̄) = 107.2 volts
The population mean (μ) = 110 volts
The sample standard deviation(s) = 47.3 volts
Number of observations (n) = 303
Standard error = SE = s /sqrt (n) = 47.3 / sqrt (303) = 2.7173 = 2.72 (approx.)
Step 3
Test statistic = (x̄ - μ) / SE = (107.2 - 110) / 2.7173 = -1.03
Assume ALPHA Level = 5% = 0.05 and DF = n-1 = 303 - 1 = 302
decision rule:since we have a left tailed test so if test statistic (found above) is less than the critical value(to be found below) then we will reject the null hypothesis.
Critical value = -1.979 (for df = 302 and alpha = 0.05 or confidence level = 0.95)
so,rejection region for null hypothesis lies left of t = -1.979 that is t < -1.979
step 4
calculated t score = -2.72 which is greater(in magnitude) than the critical value and lie left of t = -1.979 and hence fall within rejection region.
So, we must reject the null hypothesis and accept the alternate hypothesis.
conclusion :at 5% significance level, there is sufficient evidence to accept the claim that a new converter generates less than 110 volts of their old product and hence they must be redesigned.