Testing Sample Means

Distribution of Sample Means 10.2 Testing sample means • Just like the sampling distribution of sample proportions, the sampling distribution of sample means will be approximately Normal. • Meaning if we took repeated random samples of the same size, found the mean of each sample, and made a histogram, the distribution would be approximately Normal. Tests of Significance for Sample Means As sample size increases the spread of the sample distribution decreases. As sample size increase the chance of extreme values for sample means also decreases. Used to test a claim about the population mean, μ, using a sample mean, x . Conditions: the same as proportions • Same as for Proportions: 1. SRS 2. n > 30 (or we believe the population is Normally distributed) 3. pop > 10n Hypotheses: H0: μ = # < HA: μ > # ≠ New Test Statistic • t not z • We also must estimate σ with s. • Since we not only estimate μ but also σ, there is a little more error in the estimation and so the Normal model doesn’t work as well. • A new model is used called Student’s t-Model. • Student’s t-Model is symmetric and unimodal like the Normal but is more spread out. • The amount of spread is dependent on the sample size. • The larger the sample size the more Student’s tModel is like the Normal model. • We use a measurement called the degrees of freedom. df = n – 1 Formula: tdf = x −μ s n P-Value P(tdf < ___) = tcdf(-10^99, ____, df) P(tdf > ___) = tcdf(____, 10^99, df) Examples: 1. What is the probability that t is less than –1.8 if n = 10? P(t9 < -1.8) = 0.0527 2. What is the probability that t is less than –1.8 if n = 50? P(t49 < -1.8) = 0.039 3. What is the probability that t is greater than 2.86 if n = 5 P(t4 > 2.86) = 0.0230 4. What is the probability that t is greater than 2.86 if n = 90? P(t89 > 2.86) = 0.0026 Conclusion Same as for proportions: Compare P-Value to alpha. • P-Value is less than alpha, Reject HO • P-Value is greater than alpha, Fail to Reject HO Test Hypotheses: H0: μ = 130 HA: μ < 130 μ = 130; x = 126; s = 15; n = 44; df = 43 Test Statistic: t 43 = 126 − 130 15 44 = −1.7689 • Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged $126, with a standard deviation of $15. The consultant who advised the city on this project predicted that parking revenues would average $130 per day. Is there evidence that the daily fees is less than the predicted average of $130 per day? P-Value = P(t43 < -1.7689) = 0.0420 Conclusion Since the P-Value is less than alpha (0.042 < 0.05) we reject the null hypothesis We have sufficient evidence that the true mean daily fees collected is less than $130. Confidence Interval • • • • Example 3: what is t* if n = 25 and 95% confidence? s ∗ Formula: x ± tdf n Use Calculator to find t*. Program: INVT Need two numbers Example 4: what is t* if n = 50 and 95% confidence? ▫ N – Sample Size ▫ C - % Confidence • Example 2: What is t* when n = 5 and 95% confidence Example: Coffee Machine Confidence Interval • Formula: x ± tdf∗ s n ∗ t 43 = 2.017 CI : 126 ± 2.017 CI : 126 ± 4.561 Example 5: what is t* if n = 500 and 95% confidence? 15 44 (121.439,130.561) • We are 95% confident that the true mean daily fees lies between $121.44 to $130.56. A coffee vending machine dispenses coffee into a paper cup. You’re supposed to get 10 ounces of coffee, but the amount varies slightly from cup to cup. Below are the amounts measured in a random sample of 20 cups. It is believed that the amount of coffee is Normal distributed. Is there evidence that the machine is shortchanging customers? 9.9 9.6 9.7 9.8 9.7 9.8 10.1 10.0 10.0 9.8 9.9 9.9 10.1 10.0 9.6 9.5 9.9 9.5 10.2 9.9 Conditions: 1. SRS? – Stated 2. n > 30 – No but population is Normally distributed 3. pop > 10n – pop > 200 Test Run 1-Var Stats to get the summary statistics x = 9.845; s = 0.199; n = 20; df = 19 Test Statistic: Hypotheses: H0: μ = 10 HA: μ < 10 P-Value = P(t19 < -3.483) = 0.0012 Conclusion Since the P-Value is less than alpha (0.0012 < 0.05) we reject H0 We have sufficient evidence that the true mean amount of coffee is less than 10 oz. t19 = 9.845 − 10 = −3.483 0.199 20 96% Confidence Interval .199 CI : 9.845 ± 2.205 20 CI : 9.845 ± 0.098 ( 9.747,9.943 ) We are 96% confident that the true mean amount of coffee lies between 9.747 oz and 9.943 oz.