In: Statistics and Probability
In a study designed to test the effectiveness of magnets for treating back? pain, 35 patients were given a treatment with magnets and also a sham treatment without magnets. Pain was measured using a scale from 0? (no pain) to 100? (extreme pain). After given the magnet? treatments, the 35 patients had pain scores with a mean of 5.0 and a standard deviation of 2.4. After being given the sham? treatments, the 35 patients had pain scores with a mean of 4.8 and a standard deviation of 2.9. Complete parts? (a) through? (c) below. Click here to view a t distribution table. LOADING... Click here to view page 1 of the standard normal distribution table. LOADING... Click here to view page 2 of the standard normal distribution table. LOADING... a. Construct the 90?% confidence interval estimate of the mean pain score for patients given the magnet treatment. What is the confidence interval estimate of the population mean mu?? nothingless thanmuless than nothing ?(Round to one decimal place as? needed.) b. Construct the 90?% confidence interval estimate of the mean pain score for patients given the sham treatment. What is the confidence interval estimate of the population mean mu?? nothingless thanmuless than nothing ?(Round to one decimal place as? needed.) c. Compare the results. Does the treatment with magnets appear to be? effective? A. Since the confidence intervalsnbsp?overlap, it appears that the magnet treatments are more effective than the sham treatments. B. Since the confidence intervalsnbsp?overlap, it appears that the magnet treatments are no more effective than the sham treatments. C. Since the confidence intervalsnbsp do not nbsp?overlap, it appears that the magnet treatments are less effective than the sham treatments. D. Since the confidence intervalsnbsp do not nbsp?overlap, it appears that the magnet treatments are no more effective than the sham treatments. Click to select your answer(s).
Confidence Interval Magnet treatment
Confidence Interval Formula:
Note: The procedure below is used:
a) if population standard deviation is unknown and
sample size is at least 30; or
b) if sample size is less than 30 and underlying
population is approximately normal.
Step 1: Find ?/2
Level of Confidence = 90%
? = 100% - (Level of Confidence) = 10%
?/2 = 5% = 0.05
Step 2: Find t?/2
Calculate t?/2 by using t-distribution with degrees of
freedom (DF) as n - 1 = 35 - 1 = 34 and ?/2 = 0.05 as right-tailed
area and left-tailed area.
t?/2 = 1.690924 (Value can be found on excel using =
TINV(0.1,34)
Step 3: Calculate Confidence Interval
Lower Bound = x? - t?/2•(s/?n) = 5 -
(1.690924)(2.4/?35) = 4.3140
Upper Bound = x? + t?/2•(s/?n) = 5 + (1.690924)(2.4/?35)
= 5.6860
Confidence
Interval = (4.3140, 5.6860)
Confidence Interval = (4.3, 5.7) (Rounded to one decimal place)
Confidence Interval Sham treatment
Confidence Interval Formula:
Note: The procedure below is used:
a) if population standard deviation is unknown and
sample size is at least 30; or
b) if sample size is less than 30 and underlying
population is approximately normal.
Step 1: Find ?/2
Level of Confidence = 90%
? = 100% - (Level of Confidence) = 10%
?/2 = 5% = 0.05
Step 2: Find t?/2
Calculate t?/2 by using t-distribution with degrees of
freedom (DF) as n - 1 = 35 - 1 = 34 and ?/2 = 0.05 as right-tailed
area and left-tailed area.
t?/2 = 1.690924 (Value can be found on Excel using = TINV(0.1,34)
Step 3: Calculate Confidence Interval
Lower Bound = x? - t?/2•(s/?n) = 4.8 -
(1.690924)(2.9/?35) = 3.9711
Upper Bound = x? + t?/2•(s/?n) = 4.8 +
(1.690924)(2.9/?35) = 5.6289
Confidence
Interval = (3.9711, 5.6289)
Confidence Interval = (4.0, 5.6) (Rounded to one decimal place)
We can see that there is overlap in confidence interval for the two treatments. Thus, Option C and D are incorrect.
When 90% confidence intervals for the means of two independent populations don't overlap, there will indeed be a statistically significant difference between the means (at the 0.05 level ofsignificance). Thus, Option A is also incorrect.
Option B is correct