Spam filters try to sort your incoming e-mails, deciding which are real messages and which are unwanted. One method used is a point system. The filter reads each incoming e-mail and assigns points according to the sender, the subject, key words in the message, and so on. The higher the point total the more likely it is that the message is unwanted. The filter has a cutoff value for the point total; any message rated lower than that cutoff passes through to your inbox, and the rest, suspected to be spam, are diverted to the junk mailbox.
We can think of the filter's decision as a hypothesis test. The null hypothesis is that the e-mail is a real message and should go in your inbox. A high point total provides evidence that the message may be spam. When there is sufficient evidence, the filter rejects the null, classifying the message as junk. This usually works pretty well, but of course, sometimes the filter makes a mistake. Complete parts (a) through (d) below.
(a) When the filter allows spam to slip through into your inbox, what kind of error is that
(b) Which kind of error is it when a real message gets classified as junk
(c) Some filters allow you to adjust the cutoff. Suppose your filter has a default cutoff of 50 points, but you reset it to 60. What impact does this change in the cutoff value have on the chance of each type of error
(d)Is the above change in cutoff analogous to choosing a higher or lower value of α for a hypothesis test
* This question is adapted from “Business Statistics” (Sharpe et al. 3e, Pearson)
Lightbulbs
From past test records it is known that the mean lifetime of the Fillips bulbs produced is 2000 hours with a standard deviation (s) of 120 hours. The manufacturer tests a random sample of 16 light bulbs to assess the reliability of the production process with the following result (in hours).
2010 |
2010 |
1529 |
2450 |
1628 |
1976 |
1379 |
2068 |
2537 |
2687 |
2128 |
2156 |
1987 |
2020 |
1879 |
2356 |
Based on the above sample, can we say the average lifetime of Fillips lightbulbs is different from the past? Conduct the test at the significance level of 5%, using the critical value or p-value approach.
State any assumption you make in your calculations.
Notes:
The current problem is a case of testing of hypothesis regarding mean lifetime ( ) of light bulbs. The standard deviation of light bulb has been provided as hours.
Assumption: Let the random variable X denotes the lifetime of the light bulbs, where it is assumed that X follows Normal Distribution, (Park, 2015).
Step 1: The null hypothesis is constructed as,
Step 2: The alternate hypothesis is taken as, {two-tailed}
Step 3: The level of significance of the test is considered as, (5% level of significance).
Step 4: Sample mean is calculated (using MS Excel) as hours and sample standard deviation as hours.
Step 5: The test statistic of the hypothesis testing is takes as where (Standard Normal Distribution) (Sekaran, & Bougie, 2016).
The value of the test statistic is calculated as
Step 6: The p-value is calculated as the area of region where
Hence, p-value corresponding to is, (Calculated using NORMSDIST function in Excel)
Step 7: As the p-value (0.095) is greater than the significance level ( ) and the calculated test statistic is smaller than critical value of the statistic ( ), the null hypothesis is rejected at (5% level of significance).
Hence, present average lifetime of light bulbs is significantly different from the past average lifetime of 2000 hours.
Park, H. M. (2015). Hypothesis testing and statistical power of a test.
Sekaran, U., & Bougie, R. (2016). Research methods for business: A skill building approach. John Wiley & Sons.
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