A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The average thickness and the standard deviation of thickness for shift 1, based on a random sample of 40 gaskets, are 10.85 mm and 0.16 mm, respectively. The similar figures for shift 2, based on a random sample of 30 gaskets, are 10.90 mm and 0.19 mm. Let µ1-µ2 be the difference in thickness between shifts 1 and 2, and assume that the population variances are equal.
(a) Find a 95% confidence interval for µ1-µ2.
(b) Based on your answer to part a, are you convinced that the gaskets from
shift 2 are, on average, wider than those from shift 1? Why or why not?
(c) How would your answers to parts a and b change if the sample sizes
were instead 300 and 250?

a)

b) We are uncertain

c) It will change significantly

Step-by-step explanation:

a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.

Since we assume that the variances are equal, we use the pooled variance given as

,

where .

The mean difference .

The confidence interval is

b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.

c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation:

m∠abd = m∠dbc

step-by-step explanation:

given : ab bisects ∠dbc

as we know that bisection means ab divides   ∠dbc   in two equal angles named as   ∠dba and   ∠abc

so, m∠abd = m∠dbc  

because ab bisects ∠dbc.

hence , the first statement m∠abd = m∠dbc is true.