Question

The mean waiting time at the​ drive-through of a​ fast-food restaurant from the time an order is placed to the time the order is received is 85.5 85.5 seconds. A manager devises a new​ drive-through system that she she believes will decrease wait time. As a​ test, she she initiates the new system at her her restaurant and measures the wait time for 10 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts​ (a) and​ (b) below. 104.4 104.4 82.7 82.7 68.3 68.3 93.5 93.5 57.6 57.6 87.4 87.4 75.8 75.8 69.8 69.8 67.3 67.3 83.2 83.2 LOADING... Click the icon to view the table of correlation coefficient critical values. ​(a) Because the sample size is​ small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot is shown below and the sample correlation coefficient is known to be r equals = 0.988 0.988. Are the conditions for testing the hypothesis​ satisfied? ▼ Yes, No, the conditions ▼ are not are satisfied. The normal probability plot ▼ is is not linear​ enough, since the correlation coefficient is ▼ less greater than the critical value. 60 75 90 105 -2 -1 0 1 2 Time (sec) Expected z-score A normal probability plot has a horizontal axis labeled Time (seconds) from 50 to 115 in increments of 5 and a vertical axis labeled Expected z-score from negative 2 to 2 in increments of 0.5. Ten plotted points closely follow the pattern of a line that rises from left to right through (57.5, negative 1.55) and (93.5, 1). All coordinates are approximate. ​(b) Is the new system​ effective? Conduct a hypothesis test using the​ P-value approach and a level of significance of alpha equals 0.01 α=0.01. First determine the appropriate hypotheses. Upper H 0 H0​: ▼ p p mu μ sigma σ ▼ less than < greater than > equals = not equals ≠ 85.5 85.5 Upper H 1 H1​: ▼ sigma σ p p mu μ ▼ less than < greater than > equals = not equals ≠ 85.5 85.5 Find the test statistic.

Answer 1

Solution let Given data 482 68.3, 93.5, 57. 6,87-4 75.8, 69.8, 67.3, 83.2. (a) No he mdition ae ot SatisedThe namal Ponobability Coefficienk is enough Plot s Canelahion Since mot los - o2 Jiea te an the Coriba Value has Plot A natmal Pobability time erperted z - se igs 2 fnom etive expecte Vetiral Qxis lab esed Pointa Ten Plotted that is es inGe mente of o.5. fomett to nijht (1-5,1) altennative hyfethesh the Patte n tou515,eative 55) and amd b) Noc to states the null Null yPothess le: M- 85-5 Aitemativr hyfothesig Hi M4 85.5 X TX 69.8t 67. 3+ 83-2 C

I(xi-)2 S- (lo4-4-)t2771) (-3-71)t (6-71) 3-71) + (832-71) S: lo- 76. 52 (15.12 13.1901 Nou to find the test- statistic 79-85-5 3.1101 VIO 1.2G18 Jp

Nou t find he f-value f-valueTstatistis df L-515/1 . P-Value 0.coc 3 P-value Decision P-value (o0003) < d (o.01) Tefae Level af neject the nu kthesis Ht) at 0-01 we signifiance